Blog entries - Codeforces

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Author: cdkrot
Developer: ch_egor

You just needed to implement the actions described in statement.

If you solution failed, then you probably forgot to use 64-bit integers or made some small mistake.

Correct solution:

C++ code

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <string>
#include <cstring>
#include <vector>
#include <stack>
#include <queue>
#include <deque>
#include <set>
#include <map>
#include <cassert>
#include <algorithm>
#include <iomanip>
#include <ctime>
#include <cmath>
#include <bitset>

#pragma comment(linker, "/STACK:256000000")

using namespace std;

typedef long long int int64;
typedef long double double80;

const int INF = (1 << 29) + 5;
const int64 LLINF = (1ll << 59) + 5;
const int MOD = 1000 * 1000 * 1000 + 7;

const int MAX_N = 1005;

int n, x, sad;
int64 answer;

int main()
{
	//freopen("input.txt", "r", stdin);
	//freopen("output.txt", "w", stdout);

	scanf("%d %d\n", &n, &x);

	answer = x;
	sad = 0;

	int cur;
	char type;
	for (int i = 0; i < n; ++i)
	{
		scanf("%c %d\n", &type, &cur);
		if (type == '+')
		{
			answer += cur;
		}
		else if (answer >= cur)
		{
			answer -= cur;
		}
		else
		{
		    ++sad;
		}
	}

	cout << answer << " " << sad << "\n";

	fclose(stdin);
	fclose(stdout);

	return 0;
}

Python code

import sys

def main():
    n, answer = map(int, sys.stdin.readline().split())
    sad = 0

    for i in range(n):
        type_of, cur = sys.stdin.readline().split()
        cur = int(cur)
        if type_of == "+":
            answer += cur
        elif answer >= cur:
            answer -= cur
        else:
            sad += 1

    sys.stdout.write(str(answer) + " " + str(sad))

main()

Div2B Little Robber Girl's Zoo

Author: ch_egor
Developer: ch_egor

We need to sort an array with strange operations — namely, to swap elements with even and odd indices in subarray of even length. Note that we can change the 2 neighboring elements, simply doing our exchange action for subarray of length 2 containing these elements. Also, note that n ≤ 100, and it is permission to do 20 000 actions, therefore, we can write any quadratic sort, which changes the neighboring elements in each iteration (bubble sort for example).
Complexity O(n²)

C++ code

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <string>
#include <cstring>
#include <vector>
#include <stack>
#include <queue>
#include <deque>
#include <set>
#include <map>
#include <cassert>
#include <algorithm>
#include <iomanip>
#include <ctime>
#include <cmath>
#include <bitset>

#pragma comment(linker, "/STACK:256000000")

using namespace std;

typedef long long int int64;
typedef long double double80;

const int INF = (1 << 29) + 5;
const int64 LLINF = (1ll << 59) + 5;
const int MOD = 1000 * 1000 * 1000 + 7;

const int MAX_N = 105;

int n;
int arr[MAX_N];

int main()
{
	//freopen("input.txt", "r", stdin);
	//freopen("output.txt", "w", stdout);

	scanf("%d", &n);

	for (int i = 0; i < n; ++i)
	{
		scanf("%d", &arr[i]);
	}

	for (int i = n - 1; i > 0; --i)
	{
		for (int j = 0; j < i; ++j)
		{
			if (arr[j] > arr[j + 1])
			{
				swap(arr[j], arr[j + 1]);
				printf("%d %d\n", j + 1, j + 2);
			}
		}
	}

	fclose(stdin);
	fclose(stdout);

	return 0;
}

Python code

n = int(input())
arr = list(map(int, input().split()))

for i in range(n - 1, 0, -1):
    for j in range(i):
        if arr[j] > arr[j + 1]:
            arr[j], arr[j + 1] = arr[j + 1], arr[j]
            print(j + 1, j + 2)

Bonus: Is anyone able to search the shortest solution of this problem? If yes, what complexity?

Div2C Robbers' watch/Div1A Robbers' watch

Author: cdkrot
Developer: themikemikovi4

In this problem we use the septimal number system. It is a very important limitation. Let's count how many digits are showed on the watch display and call it cnt. If cnt more than 7, the answer is clearly 0 (because of pigeonhole principle). If cnt is not greater than 7, then you can just bruteforces all cases.
Depending on the implementation it will be O(BASE BASE^BASE), O(BASE^BASE) or O(BASE BASE!), where BASE = 7. Actually the most simple implementation is just to cycle between all posible hour:minute combinations and check them.
In the worst case, it will work in O(BASE BASE^BASE).

C++ code

// Copyright (C) 2016 Sayutin Dmitry.
//
// This program is free software; you can redistribute it and/or
// modify it under the terms of the GNU General Public License as
// published by the Free Software Foundation; version 3

// This program is distributed in the hope that it will be useful,
// but WITHOUT ANY WARRANTY; without even the implied warranty of
// MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the
// GNU General Public License for more details.

// You should have received a copy of the GNU General Public License
// along with this program; If not, see <http://www.gnu.org/licenses/>.


#include <iostream>
#include <vector>
#include <stdint.h>
#include <algorithm>
#include <set>
#include <map>
#include <array>
#include <queue>
#include <stack>
#include <functional>
#include <utility>
#include <string>
#include <assert.h>
#include <iterator>

using std::cin;
using std::cout;
using std::cerr;

using std::vector;
using std::map;
using std::array;
using std::set;
using std::string;

using std::pair;
using std::make_pair;

using std::min;
using std::abs;
using std::max;

using std::sort;
using std::generate;
using std::min_element;
using std::max_element;

template <typename T>
T input() {
    T res;
    cin >> res;
    return res;
}

int main() {
    std::iostream::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);

    size_t n, m;
    cin >> n >> m;

    size_t len1 = 1, len2 = 1;
    for (size_t a = 7; a < n; a *= 7)
        len1 += 1;
    for (size_t b = 7; b < m; b *= 7)
        len2 += 1;

    size_t ans = 0;
    if (len1 + len2 <= 7)
        for (size_t i = 0; i != n; ++i)
            for (size_t j = 0; j != m; ++j) {
                vector<size_t> used(7, 0);
                
                for (size_t a = i, k = 0; k != len1; a /= 7, ++k)
                    used[a % 7] += 1;
                for (size_t b = j, k = 0; k != len2; b /= 7, ++k)
                    used[b % 7] += 1;
                
                if (*std::max_element(used.begin(), used.end()) <= 1)
                    ++ans;
            }
    
    cout << ans << "\n";
    
    return 0;
}

Python code

BASE = 7

def itov(x):
    digits = []
    if x == 0:
        digits.append(0)
    while x > 0:
        digits.append(x % BASE)
        x //= BASE
    digits.reverse()
    return digits

def gen(pos = 0, minute = False, smaller = False):
    max_val = max_minute if minute else max_hour
    if pos >= len(max_val):
        if minute:
            return 1
        else:
            return gen(0, True)
    else:
        ans = 0
        for digit in range(BASE):
            if not used[digit] and (smaller or digit <= max_val[pos]):
                used[digit] = True
                ans += gen(pos + 1, minute, smaller or digit < max_val[pos])
                used[digit] = False
        return ans

n, m = map(int, input().split())
n -= 1
m -= 1
used = [False] * BASE
max_hour = itov(n)
max_minute = itov(m)
print(gen())

Bonus: Suppose the base is not fixed. Solve a problem with n ≤ 10⁹, m ≤ 10⁹, BASE ≤ 10⁹. Can you do it in $\text{[math]}$ ?

Div2D Kay and Snowflake/Div1B Kay and Snowflake

Author: cdkrot
Developers: ch_egor, cdkrot

There are many possible approaches.

Solution by cdkrot:

Look at the all candidates for the centroid of the vertices v subtree. The size of centroid subtree must be at least $\text{[math]}$ of the vertex v subtree size. (If it isn't, then after cutting the upper part will have too big size)

Choose the vertex with the smallest subtree size satisfying the constraint above. Let's prove, that this vertex is centroid indeed. If it isn't, then after cutting some part will have subtree size greater than $\text{[math]}$ of subtree size of query vertex. It isn't upper part (because of constraint above), it is one of our sons. Ouch, it's subtree less than of selected vertex, and it's still greater than $\text{[math]}$ of subtree size of query vertex. Contradiction.

So we find a centroid.
We write the euler tour of tree and we will use a 2D segment tree in order to search for a vertex quickly.
Complexity $\text{[math]}$

C++ code

// Copyright (C) 2016 Sayutin Dmitry.
//
// This program is free software; you can redistribute it and/or
// modify it under the terms of the GNU General Public License as
// published by the Free Software Foundation; version 3

// This program is distributed in the hope that it will be useful,
// but WITHOUT ANY WARRANTY; without even the implied warranty of
// MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the
// GNU General Public License for more details.

// You should have received a copy of the GNU General Public License
// along with this program; If not, see <http://www.gnu.org/licenses/>.


#include <iostream>
#include <vector>
#include <stdint.h>
#include <algorithm>
#include <set>
#include <map>
#include <array>
#include <queue>
#include <stack>
#include <functional>
#include <utility>
#include <string>
#include <assert.h>
#include <iterator>

using std::cin;
using std::cout;
using std::cerr;

using std::vector;
using std::map;
using std::array;
using std::set;
using std::string;

using std::pair;
using std::make_pair;

using std::min;
using std::abs;
using std::max;

using std::sort;
using std::generate;
using std::min_element;
using std::max_element;

template <typename T>
T input() {
    T res;
    cin >> res;
    return res;
}

const size_t MAXN = (1 << 19);
vector<size_t> graph[MAXN];

vector<pair<size_t, size_t>> segm[2 * MAXN - 1];

#define time TIME_FUCKLIBC

pair<size_t, size_t> times[MAXN];
size_t time = 0;

size_t sizes[MAXN];

void dfs(size_t v) {
    size_t sz = 1;
    
    times[v].first = time++;

    for (size_t u: graph[v]) {
        dfs(u);
        sz += sizes[u];
    }
    
    times[v].second = time;
    segm[MAXN - 1 + times[v].first].emplace_back(sz, v);
    
    sizes[v] = sz;
}

void buildseg(size_t node, size_t L, size_t R) {
    if (L != R - 1) {
        size_t mid = L + (R - L) / 2;
        buildseg(2 * node + 1, L, mid);
        buildseg(2 * node + 2, mid, R);
        
        segm[node].resize(segm[2 * node + 1].size() + segm[2 * node + 2].size());
        std::merge(segm[2 * node + 1].begin(), segm[2 * node + 1].end(), segm[2 * node + 2].begin(), segm[2 * node + 2].end(), segm[node].begin());
    }
}

pair<size_t, size_t> lowerb(size_t node, size_t L, size_t R, size_t rl, size_t rr, pair<size_t, size_t> minval) {
    if (rl <= L and R <= rr) {
        auto it = std::lower_bound(segm[node].begin(), segm[node].end(), minval);
        if (it == segm[node].end())
            return make_pair(SIZE_MAX, SIZE_MAX);
        else
            return *it;
    }
    if (R <= rl or rr <= L)
        return make_pair(SIZE_MAX, SIZE_MAX);

    size_t mid = L + (R - L) / 2;
    return min(lowerb(2 * node + 1, L, mid, rl, rr, minval)
              ,lowerb(2 * node + 2, mid, R, rl, rr, minval));
}

int main() {
    std::iostream::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);

    size_t n, q;
    cin >> n >> q;
    
    for (size_t i = 1; i != n; ++i)
        graph[input<size_t>() - 1].push_back(i);

    dfs(0);
    buildseg(0, 0, MAXN);
    
    for (size_t i = 0; i != q; ++i) {
        size_t v = input<size_t>() - 1;
        cout << lowerb(0, 0, MAXN, times[v].first, times[v].second, make_pair((sizes[v] + 1) / 2, 0)).second + 1 << "\n";
    }
    
    return 0;
}

You can consider all answers by one in dfs using this idea. Use std::set for pair (subtree size, vertex number) and at each vertex merge sets obtained from children. (Also known as "merging sets" idea)
Thus we get the answers for all vertex and we need only output answers for queries.
Complexity $\text{[math]}$

C++ code

// Copyright (C) 2016 Sayutin Dmitry.
//
// This program is free software; you can redistribute it and/or
// modify it under the terms of the GNU General Public License as
// published by the Free Software Foundation; version 3

// This program is distributed in the hope that it will be useful,
// but WITHOUT ANY WARRANTY; without even the implied warranty of
// MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the
// GNU General Public License for more details.

// You should have received a copy of the GNU General Public License
// along with this program; If not, see <http://www.gnu.org/licenses/>.


#include <iostream>
#include <vector>
#include <stdint.h>
#include <algorithm>
#include <set>
#include <map>
#include <array>
#include <queue>
#include <stack>
#include <functional>
#include <utility>
#include <string>
#include <assert.h>
#include <iterator>

using std::cin;
using std::cout;
using std::cerr;

using std::vector;
using std::map;
using std::array;
using std::set;
using std::string;

using std::pair;
using std::make_pair;

using std::min;
using std::abs;
using std::max;

using std::sort;
using std::generate;
using std::min_element;
using std::max_element;

template <typename T>
T input() {
    T res;
    cin >> res;
    return res;
}

set<pair<size_t, size_t>> solve(const vector<vector<size_t>>& graph, vector<size_t>& ans, size_t v) {
    set<pair<size_t, size_t>> cur;

    for (size_t u: graph[v]) {
        auto ch = solve(graph, ans, u);
        if (not (ch.size() < cur.size()))
            std::swap(ch, cur);
        
        for (auto elem: ch)
            cur.insert(elem);
    }

    size_t sz = cur.size() + 1;
    cur.emplace(sz, v);

    ans[v] = cur.lower_bound(make_pair((sz + 1) / 2, 0))->second;
    
    return std::move(cur);
}

int main() {
    std::iostream::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);

    size_t n, q;
    cin >> n >> q;
    
    vector<vector<size_t>> graph(n);
    for (size_t i = 1; i != n; ++i)
        graph[input<size_t>() - 1].push_back(i);

    vector<size_t> ans(n, SIZE_MAX);

    solve(graph, ans, 0);
    
    for (size_t i = 0; i != q; ++i)
        cout << ans[input<size_t>() - 1] + 1 << "\n";
    
    return 0;
}

Solution by ch_egor:
Solve it for all subtrees. We can solve the problem for some subtree, after solving the problem for all of it's children. Let's choose the heaviest child. Note that the centroid of the child after a certain lifting goes into our centroid. Let's model the lifting.
Thus we get the answers for all vertex and we need only output answers for queries.
Complexity O(n + q)

C++ code

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <string>
#include <cstring>
#include <vector>
#include <stack>
#include <queue>
#include <deque>
#include <set>
#include <map>
#include <cassert>
#include <algorithm>
#include <iomanip>
#include <ctime>
#include <cmath>
#include <bitset>

#pragma comment(linker, "/STACK:256000000")

using namespace std;

typedef long long int int64;
typedef long double double80;

const int INF = (1 << 29) + 5;
const int64 LLINF = (1ll << 59) + 5;
const int MOD = 1000 * 1000 * 1000 + 7;

const int MAX_N = 300 * 1000 + 5;
const int MAX_Q = 300 * 1000 + 5;

int n, q;
vector<int> graph[MAX_N];
int size_of[MAX_N];
int prev_of[MAX_N];
int big_subtree[MAX_N];
int centroid[MAX_N];

bool is_centroid_of_subtree(int v, int c)
{
	return ((size_of[v] - size_of[c]) * 2 <= size_of[v] && big_subtree[c] * 2 <= size_of[v]);
}

void dfs_calc(int v, int p)
{
	big_subtree[v] = 0;
	size_of[v] = 1;
	prev_of[v] = p;

	for (int i = 0; i < graph[v].size(); ++i)
	{
		dfs_calc(graph[v][i], v);
		size_of[v] += size_of[graph[v][i]];
		big_subtree[v] = max(big_subtree[v], size_of[graph[v][i]]);
	}
}

void dfs_centroid(int v)
{
	if (size_of[v] == 1)
	{
		centroid[v] = v;
	}
	else
	{
		int ptr = 0;
		for (int i = 0; i < graph[v].size(); ++i)
		{
			dfs_centroid(graph[v][i]);
			if (size_of[graph[v][ptr]] < size_of[graph[v][i]])
			{
				ptr = i;
			}
		}

		int c = centroid[graph[v][ptr]];

		while (!is_centroid_of_subtree(v, c))
		{
			c = prev_of[c];
		}

		centroid[v] = c;
	}
}

int main()
{
	//freopen("input.txt", "r", stdin);
	//freopen("output.txt", "w", stdout);

	scanf("%d %d", &n, &q);

	int v;
	for (int i = 0; i < n - 1; ++i)
	{
		scanf("%d", &v);
		graph[v - 1].push_back(i + 1);
	}

	dfs_calc(0, 0);
	dfs_centroid(0);

	for (int i = 0; i < q; ++i)
	{
		scanf("%d", &v);
		printf("%d\n", centroid[v - 1] + 1);
	}

	fclose(stdin);
	fclose(stdout);

	return 0;
}

Div2E Optimal Point/Div1C Optimal Point

Author: cdkrot
Developer: cdkrot

Let's say few words about ternary search. It works correctly, but too slow.
It's complexity is $\text{[math]}$ , where n = 10⁵, and v = 10¹⁸.
Don't use it.

Solution.
1) Let's make binary search on answer
2) Consider areas of "good" points (with dist ≤ mid) for each source point.
3) Intersect those areas and check for solution.

This area can be decsribed by following inequalities:
(You can achieve them if you expand modules in inequality "manhattan distance <= const")

.. ≤ x + y + z ≤ ..

.. ≤ - x + y + z ≤ ..

.. ≤ x - y + z ≤ ..

.. ≤ x + y - z ≤ ..

.. denote some constants.

If you intersect set of such system, you will get system of the same form, so let's learn how to solve such system.

Let's replace some variables:

a = - x + y + z

b = x - y + z

c = x + y - z

Then:

x + y + z = a + b + c

x = (b + c) / 2

y = (a + c) / 2

z = (a + b) / 2

Now the system transforms into:

.. ≤ a ≤ ..

.. ≤ b ≤ ..

.. ≤ c ≤ ..

.. ≤ a + b + c ≤ ..

We need to check if the system has solution in integers, also the numbers a, b, c must be of the same parity (have equal remainder after division by 2). (This is required for x, y, z to be integers too)

Let's get rid of parity constraint. Bruteforce the parity of a, b, c (two cases)

Make following replacement: a = 2a' + r, b = 2b' + r, c = 2c' + r, where r = 0 or r = 1. Substitute in equations above, simplify, and gain the same system for a', b', c' without parity constraint.

And now the system can be solved greedy (At first set a, b, c with minimum, and then raise slowly if necessary).

Total complexity is $\text{[math]}$ .

C++ code

// Copyright (C) 2016 Sayutin Dmitry.
//
// This program is free software; you can redistribute it and/or
// modify it under the terms of the GNU General Public License as
// published by the Free Software Foundation; version 3

// This program is distributed in the hope that it will be useful,
// but WITHOUT ANY WARRANTY; without even the implied warranty of
// MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the
// GNU General Public License for more details.

// You should have received a copy of the GNU General Public License
// along with this program; If not, see <http://www.gnu.org/licenses/>.


#include <iostream>
#include <vector>
#include <stdint.h>
#include <algorithm>
#include <set>
#include <map>
#include <array>
#include <queue>
#include <stack>
#include <functional>
#include <utility>
#include <string>
#include <assert.h>
#include <iterator>

using std::cin;
using std::cout;
using std::cerr;

using std::vector;
using std::map;
using std::array;
using std::set;
using std::string;

using std::pair;
using std::make_pair;

using std::min;
using std::abs;
using std::max;

using std::sort;
using std::generate;
using std::min_element;
using std::max_element;

template <typename T>
T input() {
    T res;
    cin >> res;
    return res;
}

struct coord {
    int64_t x;
    int64_t y;
    int64_t z;
};

struct equation {
    pair<int64_t, int64_t> S;
    pair<int64_t, int64_t> a;
    pair<int64_t, int64_t> b;
    pair<int64_t, int64_t> c;
};

vector<coord> list;

equation operator+(const equation& e1, const equation& e2) {
    equation res;
    res.S = {max(e1.S.first, e2.S.first), min(e1.S.second, e2.S.second)};
    res.a = {max(e1.a.first, e2.a.first), min(e1.a.second, e2.a.second)};
    res.b = {max(e1.b.first, e2.b.first), min(e1.b.second, e2.b.second)};
    res.c = {max(e1.c.first, e2.c.first), min(e1.c.second, e2.c.second)};
    return res;
}

coord get_solution(const equation& eq) {
    if ((eq.S.first > eq.S.second)
     or (eq.a.first > eq.a.second)
     or (eq.b.first > eq.b.second)
     or (eq.c.first > eq.c.second))
        return coord {INT64_MAX, INT64_MAX, INT64_MAX};
        
     if ((eq.a.first + eq.b.first + eq.c.first > eq.S.second)
     or (eq.a.second + eq.b.second + eq.c.second < eq.S.first))
        return coord {INT64_MAX, INT64_MAX, INT64_MAX};

     coord res;
     res.x = eq.a.first;
     res.y = eq.b.first;
     res.z = eq.c.first;
     int64_t delta = max(int64_t(0), eq.S.first - res.x - res.y - res.z);

     res.x += min(delta, eq.a.second - eq.a.first);
     delta -= min(delta, eq.a.second - eq.a.first);
     res.y += min(delta, eq.b.second - eq.b.first);
     delta -= min(delta, eq.b.second - eq.b.first);
     res.z += min(delta, eq.c.second - eq.c.first);
     delta -= min(delta, eq.c.second - eq.c.first);

     assert(delta == 0);
     
     return res;
}

int64_t DIV2(int64_t arg) {
    return (arg - (arg & 1)) / 2;
}

coord can(int64_t MAXANS) {
    equation eq;
    eq.S = eq.a = eq.b = eq.c = {INT64_MIN, INT64_MAX};
    
    for (const coord& crd: list) {
        equation nw;
        nw.S = { crd.x + crd.y + crd.z - MAXANS,  crd.x + crd.y + crd.z + MAXANS};
        nw.a = {-crd.x + crd.y + crd.z - MAXANS, -crd.x + crd.y + crd.z + MAXANS};
        nw.b = { crd.x - crd.y + crd.z - MAXANS,  crd.x - crd.y + crd.z + MAXANS};
        nw.c = { crd.x + crd.y - crd.z - MAXANS,  crd.x + crd.y - crd.z + MAXANS};
        
        eq = eq + nw;
    }

    for (int64_t r = 0; r <= 1; ++r) {
        equation tr = eq;

        tr.S.first  = DIV2(tr.S.first - 3 * r + 1);
        tr.a.first  = DIV2(tr.a.first - r + 1);
        tr.b.first  = DIV2(tr.b.first - r + 1);
        tr.c.first  = DIV2(tr.c.first - r + 1);

        tr.S.second = DIV2(tr.S.second - 3 * r);
        tr.a.second = DIV2(tr.a.second - r);
        tr.b.second = DIV2(tr.b.second - r);
        tr.c.second = DIV2(tr.c.second - r);

        coord sol = get_solution(tr);
        if (sol.x != INT64_MAX) {
            coord ans;
            ans.x = r + sol.y + sol.z;
            ans.y = r + sol.x + sol.z;
            ans.z = r + sol.x + sol.y;

            return ans;
        }
    }
    return coord {INT64_MAX, INT64_MAX, INT64_MAX};
}

int main() {
    std::iostream::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);

    list.reserve(100000);
    
    for (size_t T = input<size_t>(); T != 0; --T) {
        list.resize(input<size_t>());
        
        for (coord& crd: list)
            cin >> crd.x >> crd.y >> crd.z;

        int64_t L = -1; // definitely imposible
        int64_t R = 3 * int64_t(1000 * 1000 * 1000) * int64_t(1000 * 1000 * 1000) + 10; // definitely possible

        while (R - L > 1) {
            int64_t M = L + (R - L) / 2;
            if (can(M).x != INT64_MAX)
                R = M;
            else
                L = M;
        }
        coord ans = can(R);
        cout << ans.x << " " << ans.y << " " << ans.z << "\n";
    }
    
    return 0;
}

Bonus. This task can be solved in a lot of modifications, for example euclidian distance instead of manhattan, 2d instead of 3d, floats instead of integers. How to solve this varations?

Div1D Kay and Eternity

Authors: platypus179, Endagorion
Developer: platypus179

Let's solve this problem with scanline. Go through all rows from left to right and maintain the array in which in j index we will store the number of points in a square with bottom left coordinate (i, j), where i is current row of scanline.

This takes O(MAXCORD²) time. Note that the set of squares that contain some of the shaded points is not very large, namely — if the point has coordinates (x, y), then the set of left bottom corners of square is defined as {(a, b)|x - k + 1 < = a < = x, y - k + 1 < = b < = y}. Let's consider each point (x, y) as the 2 events:
Add one to the all elements with indexes from y - k + 1 to y on the row x - k + 1 and take one at the same interval on the row x + 1. How to calculate answer? Suppose we update the value of a cell on the row a, and before it was updated the value x on the row b. Let add to the answer for the number of squares containing x points value a - b. We can implement the addition of the segment directly and have O(nk) for processing all the events that fit in time limit. To get rid of O(MAXCORD) memory, we need to write all interested in the coordinates before processing events (them no more than nk) and reduce the coordinates in the events. It takes $\text{[math]}$ time and O(nk) memory. Now we can execute the previous point in O(nk) memory.
Complexity is $\text{[math]}$ time and O(nk) memory.

C++ code

#include <iostream>
#include <cstdio>
#include <vector>
#include <string>
#include <algorithm>

using namespace std;
typedef long long ll;

#define NotANumber -1791791791

struct query {
    int row, l, r; bool a;
    int lbl, ubr;
    query(int _row, int _l, int _r, bool _a) {row = _row; l = _l; r = _r; a = _a;}
    bool operator<(const query& other) const {
        return row < other.row;
    }
};

int main() {
    int n, k;
    scanf("%d %d", &n, &k);
    vector<pair<int, int> > points(n);
    for (int i = 0; i < n; i++)
        scanf("%d %d", &points[i].first, &points[i].second);

    sort(points.begin(), points.end());
    vector<int> xcoords;
    for (int i = 0; i < n; i++) {
        int st;
        if (xcoords.empty())
            st = points[i].first - k + 1;
        else
            st = max(xcoords.back() + 1, points[i].first - k + 1);
        for (int j = st; j <= points[i].first; j++)
            xcoords.push_back(j);
    }

    vector<query> qs;
    for (int i = 0; i < n; i++) {
        qs.push_back(query(points[i].second - k + 1, points[i].first - k + 1, points[i].first, 1));
        qs.push_back(query(points[i].second + 1, points[i].first - k + 1, points[i].first, 0));
    }

    for (query & q : qs) {
        q.lbl = lower_bound(xcoords.begin(), xcoords.end(), q.l) - xcoords.begin();
        q.ubr = upper_bound(xcoords.begin(), xcoords.end(), q.r) - xcoords.begin();
    }
    
    int xcdssz = xcoords.size();
    xcoords.clear();
    xcoords.shrink_to_fit();

    sort(qs.begin(), qs.end());

    vector<int> vec(xcdssz, 0);
    vector<int> lastchanged(xcdssz, NotANumber);
    vector<ll> ans(n);

    for (query q : qs) {
        int lbl = q.lbl;
        int ubr = q.ubr;
        for (int i = lbl; i < ubr; i++) {
            if (lastchanged[i] != NotANumber && vec[i] != 0) {
                ans[vec[i] - 1] += q.row - lastchanged[i];
            }
            if (q.a)
                vec[i]++;
            else
                vec[i]--;
            lastchanged[i] = q.row;
        }
    }

    for (int i = 0; i < n; i++)
        printf("%lld ", ans[i]);
    printf("\n");
    return 0;
}

Bonus: $\text{[math]}$ time and O(n) memory.

C++ code

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <string>
#include <cstring>
#include <vector>
#include <stack>
#include <queue>
#include <deque>
#include <set>
#include <map>
#include <cassert>
#include <algorithm>
#include <iomanip>
#include <ctime>
#include <cmath>
#include <bitset>

#pragma comment(linker, "/STACK:256000000")

using namespace std;

typedef long long int int64;
typedef long double double80;

const int INF = (1 << 29) + 5;
const int64 LLINF = (1ll << 59) + 5;
const int MOD = 1000 * 1000 * 1000 + 7;

const int MAX_N = 1000 * 100 + 5;
const int MAX_K = 305;

/** Interface */

inline int readChar();
template <class T = int> inline T readInt();
template <class T> inline void writeInt(T x);
inline void writeChar(int x);
inline void writeWord(const char *s);
inline void flush();

/** Read */

static const int buf_size = 20 * 1000 * 1000;

inline int getChar()
{
	static char buf[buf_size];
	static int len = 0, pos = 0;
	if (pos == len)
		pos = 0, len = fread(buf, 1, buf_size, stdin);
	if (pos == len)
		return -1;
	return buf[pos++];
}

inline int readChar()
{
	int c = getChar();
	while (c <= 32)
		c = getChar();
	return c;
}

template <class T>
inline T readInt()
{
	int s = 1, c = readChar();
	T x = 0;
	if (c == '-')
		s = -1, c = getChar();
	while ('0' <= c && c <= '9')
		x = x * 10 + c - '0', c = getChar();
	return s == 1 ? x : -x;
}

inline int64 readInt64()
{
	int s = 1, c = readChar();
	int64 x = 0;
	if (c == '-')
		s = -1, c = getChar();
	while ('0' <= c && c <= '9')
		x = x * 10 + c - '0', c = getChar();
	return s == 1 ? x : -x;
}

/** Write */

static int write_pos = 0;
static char write_buf[buf_size];

inline void writeChar(int x)
{
	if (write_pos == buf_size)
		fwrite(write_buf, 1, buf_size, stdout), write_pos = 0;
	write_buf[write_pos++] = x;
}

inline void flush()
{
	if (write_pos)
		fwrite(write_buf, 1, write_pos, stdout), write_pos = 0;
}

template <class T>
inline void writeInt(T x)
{
	if (x < 0)
		writeChar('-'), x = -x;

	char s[24];
	int n = 0;
	while (x || !n)
		s[n++] = '0' + x % 10, x /= 10;
	while (n--)
		writeChar(s[n]);
}

inline void writeInt64(int64 x)
{
	if (x < 0)
		writeChar('-'), x = -x;

	char s[24];
	int n = 0;
	while (x || !n)
		s[n++] = '0' + x % 10, x /= 10;
	while (n--)
		writeChar(s[n]);
}

inline void writeWord(const char *s)
{
	while (*s)
		writeChar(*s++);
}

struct point
{
	int x;
	int y;
};

struct ask
{
	int l;
	int r;
	int ptr;
	int type;
};

bool cmp(const ask &a, const ask &b)
{
	return a.ptr < b.ptr;
}

int n, k;
point arr[MAX_N];
int cord[MAX_N * 2];
ask asks[MAX_N * 2];
int cord_len = 0;
int asks_len = 0;
bool have_prev_cord[MAX_N * 2];
bool have_prev_line[MAX_N * 2];
int last_of_cord[MAX_N * 2];
int last_of_line[MAX_N * 2];
int at_cord[MAX_N * 2];
int at_line[MAX_N * 2];
int64 answer[MAX_N];

int main()
{
	//freopen("input.txt", "r", stdin);
	//freopen("output.txt", "w", stdout);

	n = readInt();
	k = readInt();

	for (int i = 0; i < n; ++i)
	{
		arr[i].x = readInt();
		arr[i].y = readInt();
		cord[cord_len++] = arr[i].x;
		cord[cord_len++] = arr[i].x - k + 1;
	}

	sort(cord, cord + 2 * n);
	cord_len = unique(cord, cord + 2 * n) - cord;
	
	for (int i = 0; i < n; ++i)
	{
		asks[asks_len].l = lower_bound(cord, cord + cord_len, arr[i].x - k + 1) - cord;
		asks[asks_len].r = lower_bound(cord, cord + cord_len, arr[i].x) - cord;
		asks[asks_len].ptr = arr[i].y - k + 1;
		asks[asks_len].type = 1;
		++asks_len;
		asks[asks_len].l = asks[asks_len - 1].l;
		asks[asks_len].r = asks[asks_len - 1].r;
		asks[asks_len].ptr = arr[i].y + 1;
		asks[asks_len].type = -1;
		++asks_len;
	}

	sort(asks, asks + asks_len, cmp);
	memset(answer, 0, sizeof(answer));
	memset(last_of_cord, 0, sizeof(last_of_cord));
	memset(last_of_line, 0, sizeof(last_of_line));
	memset(at_cord, 0, sizeof(at_cord));
	memset(at_line, 0, sizeof(at_line));
	memset(have_prev_cord, 0, sizeof(have_prev_cord));
	memset(have_prev_line, 0, sizeof(have_prev_cord));

	for (int i = 0; i < asks_len; ++i)
	{
		for (int j = asks[i].l; j <= asks[i].r; ++j)
		{
			if (!have_prev_cord[j])
			{
				have_prev_cord[j] = true;
				last_of_cord[j] = asks[i].ptr;
				at_cord[j] += asks[i].type;
			}
			else
			{
				answer[at_cord[j]] += asks[i].ptr - last_of_cord[j];

				last_of_cord[j] = asks[i].ptr;
				at_cord[j] += asks[i].type;
			}
		

			if (j != asks[i].r)
			{
				if (!have_prev_line[j])
				{
					have_prev_line[j] = true;
					last_of_line[j] = asks[i].ptr;
					at_line[j] += asks[i].type;
				}
				else
				{
					answer[at_line[j]] += 1ll * (cord[j + 1] - cord[j] - 1) * (asks[i].ptr - last_of_line[j]);

					last_of_line[j] = asks[i].ptr;
					at_line[j] += asks[i].type;
				}
			}
		}
	}

	for (int i = 1; i <= n; ++i)
	{
		writeInt64(answer[i]);
		writeChar(' ');
	}

	flush();

	fclose(stdin);
	fclose(stdout);

	return 0;
}

Div1E Travelling Through the Snow Queen's Kingdom

Authors: cdkrot, GlebsHP
Developers: ch_egor, cdkrot

Unfortunately, due to my, cdkrot, failure, it was possible to get Ac with O(nm) solution, I apologise for that.

We propose following solution:
Let's solve task with divide & conquer. At first let's lift l to the first index, where s was mentioned And lower the r to the last index, where t was mentioned. This will not affect answers, but will make implementation much more easy.

Let's look on all queries. For each query consider it's location relative to centre of edges array. If it's stricly on the left half or on the right half, then solve recursively (You need to implement function like solve(requests, l, r)).

How to answer on query, if it contains the centre? Let's precalculate two dp's:

dp_r[i] = bitset of vertices you can from to the v_i or u_i with l = m and r = i.

dp_l[i] = bitset of vertices you can go to starting from v_i or u_i with l = i and r = m - 1

v_i and u_i are vertices of i'th edge.

Using this dp the answer is yes if and ony if dp_l[l][u] = true and dp_r[r][u] = true for some u.

All above can be implemented using bitwise operations.
So the time is $\text{[math]}$

C++ code

// Copyright (C) 2016 Sayutin Dmitry.
//
// This program is free software; you can redistribute it and/or
// modify it under the terms of the GNU General Public License as
// published by the Free Software Foundation; version 3

// This program is distributed in the hope that it will be useful,
// but WITHOUT ANY WARRANTY; without even the implied warranty of
// MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the
// GNU General Public License for more details.

// You should have received a copy of the GNU General Public License
// along with this program; If not, see <http://www.gnu.org/licenses/>.


#include <iostream>
#include <vector>
#include <stdint.h>
#include <algorithm>
#include <set>
#include <map>
#include <array>
#include <queue>
#include <stack>
#include <functional>
#include <utility>
#include <string>
#include <assert.h>
#include <iterator>
#include <bitset>

using std::cin;
using std::cout;
using std::cerr;

using std::vector;
using std::map;
using std::array;
using std::set;
using std::string;

using std::pair;
using std::make_pair;

using std::min;
using std::abs;
using std::max;

using std::sort;
using std::generate;
using std::min_element;
using std::max_element;

#define size_t uint32_t

struct Req {
    size_t L;
    size_t R;
    size_t S;
    size_t T;
    size_t ID;
};

struct Edge {
    size_t v;
    size_t u;
    size_t prev1;
    size_t prev2;
    size_t next1;
    size_t next2;
};

const size_t MAXQ = 200000;
const size_t MAXN = 1000;
const size_t MAXM = 200000;

char answers[MAXQ];
Edge list[MAXM];
vector<size_t> last[MAXN];

std::bitset<MAXN> masks_r[MAXM / 2 + 10];
std::bitset<MAXN> masks_l[MAXM / 2 + 10];

void solve_layer(vector<Req>& requests, size_t l, size_t m, size_t r) {
    bool has = false;
    for (size_t i = 0; i != requests.size(); ++i)
        has or_eq (requests[i].L <= m - 1 and requests[i].R >= m);
    if (not has)
        return;
    
    for (size_t i = 0; i != r - m; ++i) {
        masks_r[i] = std::bitset<MAXN>();
        if (list[m + i].prev1 >= m and list[m + i].prev1 != std::numeric_limits<size_t>::max())
            masks_r[i] |= masks_r[list[m + i].prev1 - m];
        if (list[m + i].prev2 >= m and list[m + i].prev2 != std::numeric_limits<size_t>::max())
            masks_r[i] |= masks_r[list[m + i].prev2 - m];

        masks_r[i][list[m + i].v] = true;
        masks_r[i][list[m + i].u] = true;
    }
    
    for (size_t z = 0; z != m - l; ++z) {
        size_t i = (m - l) - 1 - z;
        masks_l[i] = std::bitset<MAXN>();
        
        if (list[l + i].next1 < m)
            masks_l[i] |= masks_l[list[l + i].next1 - l];
        if (list[l + i].next2 < m)
            masks_l[i] |= masks_l[list[l + i].next2 - l];
        masks_l[i][list[l + i].v] = true;
        masks_l[i][list[l + i].u] = true;
    }

    for (size_t i = 0; i != requests.size();)
        if (requests[i].L <= m - 1 and requests[i].R >= m) {
            answers[requests[i].ID] = (masks_r[requests[i].R - m] & masks_l[requests[i].L - l]).any();
            std::swap(requests[i], requests.back());
            requests.pop_back();
        } else {
            ++i;
        }
}

void solve(vector<Req>& requests, size_t l, size_t r) {
    if (l == r - 1) {
        for (auto& req: requests)
            answers[req.ID] = (req.S == list[l].v or req.S == list[l].u)
                          and (req.T == list[l].v or req.T == list[l].u);
        return;
    }
    size_t m = l + (r - l) / 2;
    // [l, m), [m, r).

    solve_layer(requests, l, m, r);

    vector<Req> right;
    for (size_t i = 0; i != requests.size();)
        if (requests[i].L >= m) {
            right.push_back(requests[i]);
            std::swap(requests[i], requests.back());
            requests.pop_back();
        } else {
            ++i;
        }

    requests.shrink_to_fit();
    solve(requests, l, m);
    solve(right, m, r);
}

int main() {
    size_t n, m, q;
    scanf("%d %d %d", &n, &m, &q);

    assert(n <= MAXN);
    assert(q <= MAXQ);
    assert(m <= MAXM);
    
    std::fill(answers, answers + q, 16);
    
    for (size_t i = 0; i != m; ++i) {
        scanf("%d %d", &list[i].v, &list[i].u);
        --list[i].v, --list[i].u;
        list[i].prev1 = list[i].prev2 = list[i].next1 = list[i].next2 = std::numeric_limits<size_t>::max();
        
        if (not last[list[i].v].empty()) {
            list[i].prev1 = last[list[i].v].back();
            
            if (list[last[list[i].v].back()].v == list[i].v)
                list[last[list[i].v].back()].next1 = i;
            else {
                list[last[list[i].v].back()].next2 = i;
            }
        }

        if (not last[list[i].u].empty()) {
            list[i].prev2 = last[list[i].u].back();
            
            if (list[last[list[i].u].back()].v == list[i].u)
                list[last[list[i].u].back()].next1 = i;
            else {
                list[last[list[i].u].back()].next2 = i;
            }
        }
        
        last[list[i].v].push_back(i);
        last[list[i].u].push_back(i);
    }

    vector<Req> requests;
    for (size_t i = 0; i != q; ++i) {
        Req req;
        scanf("%d %d %d %d", &req.L, &req.R, &req.S, &req.T);
        --req.L, --req.R, --req.S, --req.T;
        req.ID = i;
        
        auto it1 = std::lower_bound(last[req.S].begin(), last[req.S].end(), req.L);
        auto it2 = std::upper_bound(last[req.T].begin(), last[req.T].end(), req.R);

        if (it1 == last[req.S].end() or it2 == last[req.T].begin())
            answers[i] = false;
        else {
            req.L = *it1;
            req.R = *std::prev(it2);
            if (req.L > req.R)
                answers[i] = false;
            else
                requests.push_back(req);
        }
    }
    
    solve(requests, 0, m);

    for (size_t i = 0; i != q; ++i) {
        printf("%s", (answers[i] ? "Yes\n" : "No\n"));
    }
    
    return 0;
}

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