Codeforces Round 966 (Div. 3) Editorial

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2000A - Primary Task

Tutorial

Solution

for _ in range(int(input())):
    a = int(input())
    if 102 <= a <= 109 or 1010 <= a <= 1099:
        print("YES")
    else:
        print("NO")

2000B - Seating in a Bus

Idea: myav

Tutorial

Solution

#include <bits/stdc++.h>
using namespace std;
void solve(){
    int n;
    cin >> n;
    vector<int>a(n);
    for(auto &i : a) cin >> i;
    int left = a[0], right = a[0];
    for(int i = 1; i < n; i++){
        if(a[i] + 1 == left) left = a[i];
        else if(a[i] - 1 == right) right = a[i];
        else {
            cout << "NO\n";
            return;
        }
    }
    cout << "YES\n";
}
 
int main(){
    int t;
    cin >> t;
    while (t--) {
        solve();
    }
    return 0;
}

2000C - Numeric String Template

Idea: myav

Tutorial

Solution

def solve():
    n = int(input())
    a = list(map(int, input().split()))
    m = int(input())
    for _ in range(m):
        m_1 = {}
        m_2 = {}
        s = input().strip()
        if len(s) != n:
            print("NO")
            continue
 
        ok = True
        for j in range(n):
            if s[j] not in m_1 and a[j] not in m_2:
                m_1[s[j]] = a[j]
                m_2[a[j]] = s[j]
            elif (s[j] in m_1 and m_1[s[j]] != a[j]) or (a[j] in m_2 and m_2[a[j]] != s[j]):
                ok = False
                break
        print("YES" if ok else "NO")
 
t = int(input())
for _ in range(t):
    solve()

2000D - Right Left Wrong

Idea: Vladosiya

Tutorial

Solution

def solve():
    n = int(input())
    a = [0]
    for x in input().split():
        x = int(x)
        a.append(a[-1] + x)
    s = input()
    ans = 0
    l = 0
    r = n - 1
    while r > l:
        while l < n and s[l] == 'R':
            l += 1
        while r >= 0 and s[r] == 'L':
            r -= 1
        if l < r:
            ans += a[r + 1] - a[l]
            l += 1
            r -= 1
    print(ans)
 
t = int(input())
for _ in range(t):
    solve()

2000E - Photoshoot for Gorillas

Idea: Vladosiya

Tutorial

Solution

#include <bits/stdc++.h>
 
#define int long long
 
using namespace std;
 
#define MAXW 200200
#define MAXNM 200200
 
int n, m, k, w, p;
int arr[MAXW], prr[MAXNM];
 
static inline int calcc(int i, int j) {
    int upr = min(i, n - k);
    int leftr = min(j, m - k);
    int upl = max(-1LL, i - k);
    int leftl = max(-1LL, j - k);
    return (upr - upl) * (leftr - leftl);
}
 
void build() {
    sort(arr, arr + w);
    reverse(arr, arr + w);
    p = 0;
    for (int i = 0; i < n; ++i)
        for (int j = 0; j < m; ++j)
            prr[p++] = calcc(i, j);
    sort(prr, prr + p);
    reverse(prr, prr + p);
}
 
int solve() {
    int sum = 0;
    for (int i = 0; i < w; ++i)
        sum += prr[i] * arr[i];
    return sum;
}
 
signed main() {
    int t; cin >> t;
    while (t--) {
        cin >> n >> m >> k >> w;
        for (int i = 0; i < w; ++i)
            cin >> arr[i];
        build();
        cout << solve() << endl;
    }
}

2000F - Color Rows and Columns

Idea: senjougaharin

Tutorial

Solution

#include <iostream>
#include <algorithm>
#include <vector>
 
using namespace std;
 
void solve() {
    int n, k;
    cin >> n >> k;
    vector<int> a(n), b(n);
    for (int i = 0; i < n; ++i) {
        cin >> a[i] >> b[i];
    }
    vector<vector<int>> dp(n + 1, vector<int>(k + 1, 1e9));
    dp[0][0] = 0;
    for (int i = 0; i < n; ++i) {
        int x = a[i], y = b[i];
        int xy = x + y;
        int cost = 0;
        for (int j = 0; j <= xy; ++j) {
            for (int j1 = 0; j1 + j <= k; ++j1) {
                dp[i + 1][j1 + j] = min(dp[i + 1][j1 + j], dp[i][j1] + cost);
            }
            if (j < xy) {
                if (x >= y) {
                    x--, cost += y;
                } else {
                    y--, cost += x;
                }
            }
        }
    }
    cout << (dp[n][k] == 1e9 ? -1 : dp[n][k]) << '\n';
}
 
int main() {
    int t;
    cin >> t;
    for (int i = 0; i < t; ++i) {
        solve();
    }
    return 0;
}

2000G - Call During the Journey

Idea: senjougaharin

Tutorial

Solution

#include <iostream>
#include <vector>
#include <set>
 
using namespace std;
 
void solve() {
    int n, m;
    cin >> n >> m;
    int t0, t1, t2;
    cin >> t0 >> t1 >> t2;
    vector<vector<vector<int>>> g(n);
    for (int i = 0; i < m; ++i) {
        int u, v, l1, l2;
        cin >> u >> v >> l1 >> l2;
        u--, v--;
        g[u].push_back({v, l1, l2});
        g[v].push_back({u, l1, l2});
    }
    set<pair<int, int>> prq;
    prq.insert({t0, n - 1});
    for (int i = 0; i + 1 < n; ++i) {
        prq.insert({-1e9, i});
    }
    vector<int> dist(n, -1e9);
    dist[n - 1] = t0;
    while (!prq.empty()) {
        auto p = *prq.rbegin();
        prq.erase(p);
        int d = p.first, u = p.second;
        for (auto e: g[u]) {
            int v = e[0], l1 = e[1], l2 = e[2];
            int d1 = (d - l1 >= t2 || d <= t1) ? d - l1 : d - l2;
            if(d1 == d - l2) d1 = max(d1, t1 - l1);
            if (dist[v] < d1) {
                prq.erase({dist[v], v});
                dist[v] = d1;
                prq.insert({d1, v});
            }
        }
    }
    cout << (dist[0] >= 0 ? dist[0] : -1) << '\n';
}
 
int main() {
    int t;
    cin >> t;
    for (int i = 0; i < t; ++i) {
        solve();
    }
    return 0;
}

2000H - Ksyusha and the Loaded Set

Idea: senjougaharin

Tutorial

Solution

#pragma GCC optimize("O3,unroll-loops")
 
#include <bits/stdc++.h>
 
#define endl '\n'
 
using namespace std;
 
typedef pair<int, int> ipair;
 
#define INF 1'000'000'009
#define MAXN 200200
#define MAXMEM 5'000'000
#define MAXLEN 2'000'100
 
#define X first
#define Y second
 
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
 
struct node {
    node *lv = nullptr, *rv = nullptr;
    ipair key;
    int prior;
    int minl = INF;
    
    node(const ipair& key) : key(key), prior(rng()), minl(key.Y) {}
};
 
static inline int minl(node *v) {
    return (v == nullptr ? INF : v->minl);
}
 
static inline void upd(node *v) {
    v->minl = min(v->key.Y, min(minl(v->lv), minl(v->rv)));
}
 
node *merge(node *s1, node *s2) {
    if (s1 == nullptr) return s2;
    if (s2 == nullptr) return s1;
    if (s1->prior > s2->prior) {
        s1->rv = merge(s1->rv, s2);
        upd(s1);
        return s1;
    } else {
        s2->lv = merge(s1, s2->lv);
        upd(s2);
        return s2;
    }
}
 
pair<node*, node*> split(node* &v, ipair x) {
    if (v == nullptr) {
        return {nullptr, nullptr};
    }
    if (v->key < x) {
        auto [s1, s2] = split(v->rv, x);
        v->rv = s1;
        upd(v);
        return {v, s2};
    } else {
        auto [s1, s2] = split(v->lv, x);
        v->lv = s2;
        upd(v);
        return {s1, v};
    }
}
 
int n, m;
int arr[MAXN];
node *mem = (node*)calloc(MAXMEM, sizeof(*mem));
node *mpos = mem;
node *root;
set<int> M;
 
// [l..r)
void add_seg(int l, int r) {
    if (l >= r) return;
    ipair p {r - l, l};
    auto [s1, s2] = split(root, p);
    node *v = mpos++;
    *v = node(p);
    root = merge(merge(s1, v), s2);
}
 
void rem_dek(node* &v, const ipair& x) {
    if (v == nullptr) return;
    if (v->key == x) {
        v = merge(v->lv, v->rv);
        if (v) upd(v);
        return;
    }
    if (v->key < x) rem_dek(v->rv, x);
    else rem_dek(v->lv, x);
    upd(v);
}
 
// [l..r)
void rem_seg(int l, int r) {
    if (l >= r) return;
    ipair p {r - l, l};
    rem_dek(root, p);
}
 
void add_val(int x) {
    auto it = M.lower_bound(x);
    int clsl = (it == M.begin() ? 0 : *prev(it));
    int clsr = (it == M.end() ? INF : *it);
    rem_seg(clsl + 1, clsr);
    add_seg(clsl + 1, x);
    if (clsr != INF) add_seg(x + 1, clsr);
    M.insert(x);
}
 
void rem_val(int x) {
    auto it = M.lower_bound(x);
    int clsl = (it == M.begin() ? 0 : *prev(it));
    int clsr = (next(it) == M.end() ? INF : *next(it));
    rem_seg(clsl + 1, x);
    rem_seg(x + 1, clsr);
    if (clsr != INF) add_seg(clsl + 1, clsr);
    M.erase(x);
}
 
int query(int k) {
    if (M.empty()) return 1;
    auto [s1, s2] = split(root, make_pair(k, -1));
    fprintf(stderr, "AMOGUS: %p, %p\n", s1, s2);
    int ans = min(*M.rbegin() + 1, minl(s2));
    root = merge(s1, s2);
    return ans;
}
 
void init() {
    root = nullptr;
    mpos = mem;
    M = {arr, arr + n};
    add_seg(1, *M.begin());
    for (auto it = M.begin(); next(it) != M.end(); ++it)
        add_seg(*it + 1, *next(it));
}
 
signed main() {
    int t; cin >> t;
    while (t--) {
        cin >> n;
        for (int i = 0; i < n; ++i)
            cin >> arr[i];
        init();
        cin >> m;
        for (int j = 0; j < m; ++j) {
            char tp; cin >> tp;
            int x, k;
            switch (tp) {
            case '+':
                cin >> x;
                add_val(x);
                break;
            case '-':
                cin >> x;
                rem_val(x);
                break;
            case '?':
                cin >> k;
                cout << query(k) << " ";
                break;
            }
        }
        cout << endl;
    }
}

auto fitr = que.lower_bound(x); if(fitr == que.end()) cout << -1 << " "; else { int mini = INT_MAX; for(; fitr != que.end();) { mini = min(mini, *(fitr->second.begin())); fitr = next(fitr); } cout << mini << " "; }

#include <bits/stdc++.h> using namespace std; // Include necessary header files for ordered_set #include <ext/pb_ds/assoc_container.hpp> #include <ext/pb_ds/tree_policy.hpp> using namespace __gnu_pbds; // Define an ordered set typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> ordered_set; // Define macros for convenience #define int long long #define f(i, a, b) for (int i = a; i < b; i++) #define rf(i, a, b) for (int i = a; i >= b; i--) #define vi vector<int> #define vpii vector<pair<int, int>> #define vvvi vector<vector<vector<int>>> #define vvi vector<vector<int>> #define vvpi vector<vector<pair<int, int>>> #define vb vector<bool> const int inf = 1e18; int n, m, t0, t1, t2; vvvi adj; int dijkstra(int src, int dest, int start_time) { vi dist(n + 1, inf); dist[src] = start_time; priority_queue<vi, vvi, greater<vi>> pq; pq.push({start_time, src}); int dur = t2 - t1; while (!pq.empty()) { vi curr = pq.top(); pq.pop(); int u = curr[1]; int time = curr[0]; if (time > dist[u]) continue; for (vi edge : adj[u]) { int v = edge[0]; int l1 = edge[1]; int l2 = edge[2]; int new_time = time + l1; if (time >= t1 && time < t2) { // Call is active if (dur >= l2) { // take l2 route new_time = time + l2; dur -= l2; } else { // take l2 route if (l2 < (t2 - time) + dur + l1) { new_time = time + l2; dur = 0; } else { // wait and take l1 route new_time = time + (t2 - time) + dur + l1; dur = 0; } } } else if (time <= t1 && t1 < new_time) { // Crosses into active call period if (dur >= l2) { // take l2 route new_time = time + l2; dur -= l2; } else { if (l2 < (t1 - time) + dur + l1) { // take l2 route new_time = time + l2; dur = 0; } else { // wait and take l1 route new_time = time + (t1 - time) + dur + l1; dur = 0; } } } if (new_time < dist[v]) { // Relaxation dist[v] = new_time; pq.push({new_time, v}); } } } return dist[dest]; } void solve() { cin >> n >> m; cin >> t0 >> t1 >> t2; adj = vvvi(n + 1); f(i, 0, m) { int u, v, l1, l2; cin >> u >> v >> l1 >> l2; adj[u].push_back({v, l1, l2}); adj[v].push_back({u, l1, l2}); } int low = 0, high = t0; int ans = -1; // binary on how late we can start our journey to reach on time while (low <= high) { int mid = low + (high - low) / 2; if (dijkstra(1, n, mid) <= t0) { ans = mid; low = mid + 1; } else { high = mid - 1; } } cout << ans << endl; } signed main() { ios_base::sync_with_stdio(false); cin.tie(NULL); int tc; cin >> tc; while (tc--) solve(); return 0; }

Comments (72)

Write comment?

Manish_Jangir

3 months ago, # |

can anyone pls tell why this code is getting a runtime error. for Problem D

Code

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define f(i,a,b) for(ll i=a;i<b;i++)
#define rf(i,a,b) for(ll i=b-1;i>=a;i--)

/* When a man learns to love he must bear the risk of hatred */
/* I don't know WTF am I doing */
/* Now I hate You */

int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(0);cout.tie(0);
    ll t;cin>>t;
    while(t--){
        ll n;cin>>n;
        vector<ll> a(n);
        f(i,0,n){
            cin>>a[i];
        }
        string str; cin>>str;
        vector<ll> leftsig,rightsig;
        f(i,0,n){
            if(str[i]=='L')leftsig.push_back(i);
            else rightsig.push_back(i);
        }
        ll left = 0;
        ll right = rightsig.size()-1;
        ll sum = 0;
        
        while(left<leftsig.size() && right>=0 && leftsig[left]<rightsig[right]){
            sum += accumulate(a.begin()+leftsig[left],a.begin()+rightsig[right]+1,0LL);
            left++;
            right--;
        }
        cout<<sum<<endl;

    }   
    
}

And also the issue with the algo pls.

→ Reply

towley

3 months ago, # ^ |

← Rev. 2 →

consider the case where there are no 'R' in the string. in this scenario, your 'right' variable would be initialized with the value -1

edit : avoid using C++17 (GCC 7-32) and use C++20 (GCC 13-64) instead

you mean something like this -> n = 6, a(n) = 1 1 1 1 1 1 s = LLLLLL -> and the output should be 0 right. actually I checked this one already and it gives me 0 perfectly fine. Cause I'm checking the out of bound condition before using those

I admit I do not know much about C++17 (32-bit), but if I had to guess, this particular error might be occurring because your while loop condition:

while(left < leftsig.size() && right >= 0 && leftsig[left] < rightsig[right]) {

is not stopping after right >= 0 evaluates to false. Instead, it continues evaluating leftsig[left] < rightsig[right], which could cause an access violation if right is -1 and thus result in the error

you can fix it by choosing C++20 (GCC 13-64) as your compiler when you submit the code

https://codeforces.net/contest/2000/submission/276452095

yeah I guess C++17 doesn't stop when a condition is false in the loop conditions, untill it checks all the conditions

edit: I used the prefix sum to avoid tle and it went ohk

oh thanks for the help it went perfectly just by changing this. Wasn't expecting it to be this sort of issue, will use C++ 20 always :) .

TANMAY_1408

This solution will give tle as you are calculating sum of subarray again and again. Instead use prefix sum to get the sum of the subarray.

I did that that in the updated code here ->

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define f(i,a,b) for(ll i=a;i<b;i++)
#define rf(i,a,b) for(ll i=b-1;i>=a;i--)

/* When a man learns to love he must bear the risk of hatred */
/* I don't know WTF am I doing */
/* Now I hate You */

int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(0);cout.tie(0);
    ll t;cin>>t;
    while(t--){
        ll n;cin>>n;
        vector<ll> a(n);
        f(i,0,n){
            cin>>a[i];
        }
        string str; cin>>str;

        vector<ll> prefix_sum(n+1, 0);
        for(ll i = 0; i < n; i++) {
            prefix_sum[i+1] = prefix_sum[i] + a[i];
        }

        vector<ll> leftsig,rightsig;
        f(i,0,n){
            if(str[i]=='L')leftsig.push_back(i);
            else rightsig.push_back(i);
        }
        ll left = 0;
        ll right = rightsig.size()-1;
        ll sum = 0;
        
        while(left<leftsig.size() && right>=0 && leftsig[left]<rightsig[right]){
            sum += prefix_sum[rightsig[right]+1] - prefix_sum[leftsig[left]];
            left++;
            right--;
        }
        cout<<sum<<endl;

    }   
    
}

But this still gives me runtime error edit : after changing from C++17 to C++20 this works perfectly fine and thanks for the prefix sum intuition

ziad-abdou1

I need help

Can anyone tell me why my solution in proplem E got TLE when i write it in java , but when i write the same code in c++ i got AC

java solution --> Your text to link here...

c++ solution --> Your text to link here...

yer_06

С++ is faster than java

WOWWA

someone explain how to intuit the way they count it in problem E please.

racel

consider 2D adjacent difference it could be a lot easier

i_would_win

do u know diffrence array? to update values in range for 1d vector it can be used here to count for 2d vector then 2d prefix sum

arvindf232

← Rev. 3 →

+17

I am impressed by how treap is being casually used in a div3 contest editorial (problem H).

(Context: Of course I am well aware that this is not how most contestants solved, or are supposed to solve it, but rather make use of the (non-essential) constraints max <= 2e6. It is also clear that a fully online solution of this without the extra constraints will have to involve your favourite BBST. Though, I don't know how many people will get confused by the editorial. )

Dominater069

+33

They took the effort to explain the std::set part but not the treap part :)

yellow_13

If possible can you explain how to do it without treap? I was thinking of a solution where I can maintain an array where the indices represent the lengths of the free segments and the value at that index is the value of the left border of that segment, then I just need to find suffix minimums to answer any query. I was thinking of using segment tree for handling updates and suffix range queries, but I think that will TL because there is no bound on sum of $$$max(a_i)$$$ across all test cases and my time complexity would become $$$O(T\cdot max(a_i) \cdot\log{max(a_i)})$$$ in that case.

You can reuse the same segment tree. So initialise one segment tree with size max(a_i), at the end of each test case you undo all insertions.

I take that you already know how to "find first index in segment tree that is >= some value", but for others this can be handled by using a max-segment tree and a standard segment tree descend

Wry_0313

I tried implementing this solution but I received TLE. Can you help me find out why this is happening? https://codeforces.net/contest/2000/submission/278601400

Your segment tree descent code is O(n). Read the segment tree before deciding going left or right, rather than just recur/

Hi I just reimplemented a log(n) descent. However, I am still getting TLE ;-;

https://codeforces.net/contest/2000/submission/278684047

I cannot find an obvious mistake. Try to run it in gym with more TL and see how close it is. If it is close to 3s, then it may simply be that recursive seg tree are too slow to pass. My solution that takes 1500ms in Kotlin is iterative.

noobcoder180

treap sounds like something you cover a pool with

lrvideckis

for E, I completely missed the n*m<=2e5 bound and coded a O(nlogn + mlogm + wlog(n*m)) solution

276188663

OmarAboutaleb77

I also misread the n*m<=2e5 constraint, and I tried to code a bfs type solution starting from the center of the grid where we greedily visit cells in the graph that have the most subarrays containing them. I failed in the implementation, but is your solution idea similar?

yes, similar, that's pretty much what I did

Red0

I got scared when I didn't see the n*m <= 2e5 condition and I immediately re-checked the problem statement :)

bkchodi

🆗

shauncodes

2 months ago, # ^ |

Thank you so much for the solution! Really helpful to learn a new approach...

I thought of BFS, but could not figure out how, but this is amazing!

Harshvardhan_Acharya

H can be solved using segment tree. 276462150

stemdarrenyang

How can I report suspected cheating in this round?

tapasyadimree

please help me with my submissions for C. I've used a map for int to char mapping, and then a vector of size 26 for char to in mapping. My conditions are also correct (acc to me) so please if anyone can see why is my code failing please help me ! I am getting WA on test 18.

276216688

so i just changed my code to use 2 maps instead of 1 map + 1 vector and it got accepted. why did this happen ?

276481407

i solved it.. i was initializing vector with -1 which is wrong. i made it INT_MIN and it worked

276482552

Oleja123

The problem H has a beautiful solution using sqrt decomposition (It works slower than the author's solution, but it seems to me easier to understand and write). My example solution: https://codeforces.net/contest/2000/submission/276433907

harshith_04

Problem H Solution without any specific method/data structure but only logic — 276496494

I_love_GSNQK

what is the time complexity of this block?

Worst case is ~ $$$\sqrt {2 . 10^6}$$$. When the set has 2e6-1, 2e6-4, 2e6-8, 2e6-13, .... and the query is "? 1" ig

SinisterBrute

For problem D, I know that it can be solved using 2 pointers or manually matching the L's with R's, but still I can't figure out what is wrong with my solution, it gives Wrong Answer on Test 7, i.e. it is working for smaller inputs but not for larger ones, (maybe, I don't know). It would be very helpful, if someone could point out the idea that I am missing.

My Submission: https://codeforces.net/contest/2000/submission/276453840

ckrian__223

Can someone explain me Problem F in detail? I saw shayan's vid but I keep getting lost. (I have a fair experience with dp)

olegmak

The hardest part of this problem is understanding optimal way of painting single rectangle. Like they say in editorial you must always choose to paint single column or row of minimum size. For example, if you have 3 by 4 rectangle you unpainted rectangle will change like this: 3x4 -> 3x3 -> 3x2 (or 2x3, does not matter) -> 2x2 -> 2x1 (or 1x2) -> 1x1 -> 0x0 Every transition will give you one score point (except for last which gives 2 points because we paint one row and one column at the same time). You perform this sequence of operations on the first rectangle, keep track on the number of operations (cnt1) and score (s), calculate recursively min. number of operations for the rest of the rectangles to achieve (k — s) score (cnt2), choose min(cnt1 + cnt2) after painting sequence is complete.

Thanks! :)

b00s

can someone explain, why the sample testcase 1 answer is 0. I think we can wait upto 19 minutes and then take the bus route 1 — 5 which will cost 30 minutes, and we will arrive on 5 at 49 minutes ?

You cannot ride a bus during entire phone call, not just start of it.

-2.71

-17

My Insights for A,B,C

Problem A

Problem B

Problem C

$$$\text{Nice Educational Contest overall !}$$$

24x7coldd

Can anyone help me find out the test case for which 276261336 problem C (test case 18), failed ?. Appreciate any help thank you :)

27debowskijakub

I found this test case:

Test

1
3
0 0 1 
4
CBB
ACE
D
DD

which in my code returns:

Result

NO
NO
NO
NO

and in yours:

NO
YES
NO
NO

Hope it helps!

This is awesome, Thanks a lot :)

wsaleem

← Rev. 4 →

For problem G sample 1, can someone share the route which corresponds to a starting time of 0? From what I understand of the problem, there should be no solution and the answer should be -1. Maybe I am overlooking something. I am copying the sample below for convenience.

joaozao

In time 0, walk through the edge from 1 to 5. That will take 100 units of time, which fits in t0.

For problem E, the image in the Notes section loaded for me without any gorillas and I lost a lot of time trying to make sense of the image.

For Problem F, sample 6, reproduced below, can someone explain how the required 25 points can be obtained in 80 operations, as given in the sample solution?

akmeena5726

First we will completely fill the 2*9 and 4*3 rectangles. After this our score would be 18 and operations are 30. After this we will start filling the 8*10 rectangle in this order — 8, 8, 8, 7, 7, 6, 6. This would lead to a total score of 25 in 80 operations.

TanJWcode

Take all 9x2 for 11 points, take all 4x3 for 7 points, take all but a 5x6 rectangle from 8x10 (which is 50 squares) for 7 points. 8+8 + 8+7 + 7+6 + 6 = 50 11+7+7 =25, 18+12+50=80

abishekiitb

are there public solutions using python? as i am currently learning python and want to master it before starting c++

___Abc123___

did any one solved F using greedy?

rk0822

Can you tell me how 35 operations are required in last test case

4 18 5 4 8 5 8 3 6 2

Use entire first and last rectangle and then use rectangle which is having b = 3 to make the score 18

Can anyone please explain what's wrong in my approach of problem G. I applied binary search to find how late we can start to reach on time. What I'm doing in my modified dijkstra is that if I can take bus then then will take it but if I will get a call in between or currently on a call then I will take the best of wait till the call is finish and take bus or walk.

My Code

Is there anyone who solved problem G using python? My code exceeded time limit on test 17.

for test in range(int(input())):
    n, m = [int(i) for i in input().split()]
    t0, t1, t2 = [int(i) for i in input().split()]

    G = dict()
    for i in range(m): #made graph
        start, end, bus, walk = [int(i) for i in input().split()]
        if start not in G:
            G[start] = [[end,(bus,walk)]]
            
        else:
            G[start].append([end,(bus,walk)])
            
        if end not in G:
            G[end] = [[start,(bus,walk)]]
        else:
            G[end].append([start,(bus,walk)])

    #Djikstras from the end (node n) to start (node 1) using latest, latest[n]=t0
    #if he can ride bus (not on call), ride bus
    #else choose max (starting time) of walking and waiting for call to end then ride bus

    latest = [-1]*(n)+[t0] 
    cur = n; vis = []; frontier = [n]

    while True: #if cur=k, then we have the shortest path to k
        cur = frontier[0] #pick largest time
        for i in frontier:
            if latest[i] > latest[cur]:
                cur = i
        if cur == 1:
            break
        frontier.remove(cur)
        vis.append(cur)

        for prev, time in G[cur]:
            if prev in vis:
                continue
            bus = time[0]; walk = time[1] 
            a = latest[cur]
            if (t2 <= a - bus) or (t1 >= a): #can ride bus
                latest[prev] = max(latest[prev], a - bus)
            else:
                latest[prev] = max(latest[prev], t1 - bus, a - walk)
            
            if prev not in frontier:
                frontier.append(prev)
        
    print(latest[1])

Nikita_Malykh

Thanks for the editorial ^_^

Can someone please explain me how are we getting 35 operations in last test case of sample? I am getting 36 operations to score 18 points.

Fill the 6 x 2 (score = 8 & operations = 12) rectangle and 5 x 4 rectangle (score = 9 & operations = 20) completely and one row of 8 x 3 rectangle (score = 1 & operations = 3). Therefore total operations and score would be 12 + 20 + 3 = 35 and 8 + 9 + 1 = 18 respectively.

Ahmed_Ashraf_f

Why am I not official in this round !!!

soicodoc10tin

In problem G, why did I get TLE when I used priority_queue but after switching to set<pair<int,int>> it became AC?

freshwater_123

I am only into dsa, can that help me in cp. I am new to this field

_L0giCAL_

2 months ago, # |

In problem H, I am getting WA, Can someone explain what am I doing wrong.

I am trying to maintain the free segments and updation using multiset. I am maintaining (dist, left_endpoint) in the multiset and the points currently in the set. Then I am performing the insertion and removal operation by taking the next and previous points from the points set.

Link to my solution: submission

kuangbiaopilihu

I did the same thing with you, and also got WA in test case 2.

Were you able to solve the issue??

I can not, but I do not know why. It seems quite right to me.

I have a test case totay and my code gets the wrong answer. Now I find the problem in my code with this case. You can test yours. input: 1 6 1 4 6 7 8 9 3 ? 5 + 10 ? 1 output 10 2

Yes my code is also failing this TC. The issue is that since I have sorted the multiset based on distance so there might be a smaller point later on but with a bigger distance. I need to find the lowest value available in suffix of distances starting from k.

Pannka384

is it possible to solve problem H using a segment tree

ha5750

Iam getting time limit exceeded for the solution for problem e, where my big O time is the same as solution.. Can anyone tell my why?283002155

qinziao

4 weeks ago, # |

Love problem F and author's beautiful and flawless solution to the problem.Back to my primitive idea,I let dp[i] be the minimum cost to reach score k,and attempt to solve the problem only within greedy and complexly loops.

Vladosiya's blog

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