Duarte's blog

Help problem SegmentTree with lazy
By Duarte, history, 9 years ago, In English

Hello everyone, I'm trying to solve the problem http://www.spoj.com/problems/HORRIBLE/, I solved using BIT, but I'm learning about Lazy Propagation and I want to solve using this. I did a code but it is getting WA, and I don't know why. Anyone can help me ?


#include <bits/stdc++.h>

using namespace std;

typedef long long int lli;
typedef vector<lli> vl;

class SegmentTree
{
private:
	vl st, lazy, A;
	int n;
	int left(int p) { return p << 1; }
	int right(int p) { return (p << 1) + 1; }	

	lli rsq(int p, int L, int R, int i, int j)
	{
		
		if(lazy[p] != 0)
		{
			st[p] += (R - L + 1) * lazy[p];  //RMQ = st[p] += lazy[
			if(L != R)
			{
				lazy[left(p)] += lazy[p];
				lazy[right(p)] += lazy[p];
			}
			lazy[p] = 0;
		}

		if(i > R || j < L || R < L) return 0LL;
		if(L >= i && R <= j) return st[p];

		return rsq(left(p), L, (L + R) / 2, i, j) + rsq(right(p), (L + R) / 2 + 1, R, i, j);
	}

	void updateRange(int p, int L, int R, int i, int j, int newValue)
	{
		if(lazy[p] != 0)
		{
			st[p] += (R - L + 1) * lazy[p];  //RMQ = st[p] += lazy[p];

			if(L != R)
			{
				lazy[left(p)] += lazy[p];
				lazy[right(p)] += lazy[p];
			}
			lazy[p] = 0;
		}

		if(L > R || L > j || R < i)
			return;
		if(L >= i && R <= j)
		{
			st[p] += (R - L + 1) * newValue; //RMQ = st[p] += value;
			if(L != R)
			{
				lazy[left(p)] += newValue;
				lazy[right(p)] += newValue;
			}
			return;
		}

		updateRange(left(p), L, (L + R) / 2, i, j, newValue);
		updateRange(right(p), (L + R) / 2 + 1, R, i, j, newValue);

		st[p] = st[left(p)] + st[right(p)];

	}



public:
	SegmentTree(int _n)
	{
		n = _n;
		st.assign(4 * n, 0);
		lazy.assign(4 * n, 0);
	}

	lli rsq(int i, int j)
	{
		return rsq(1, 0, n - 1, i, j);
	}
	void updateRange(int i, int j, int newValue)
	{
		updateRange(1, 0, n - 1, i, j, newValue);
	}
};

int main()
{
	int T;
	scanf("%d", &T);
	while(T--)
	{
		int n, c;
		scanf("%d %d", &n, &c);
		SegmentTree st(n);

		int x, p, q, v;
		while(c--)
		{
			scanf("%d %d %d", &x, &p, &q);
			if(x == 1)
				printf("%lld\n", st.rsq(p - 1, q - 1));
			else
			{
				scanf("%d", &v);
				st.updateRange(p - 1, q - 1, v);
			}
		}
	}
	return 0;
}
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Comments Comments (7)
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9 years ago, # |
← Rev. 3   Vote: I like it 0 Vote: I do not like it

lazy update is wrong. it should be like this : lazy[L(p)] += value*(mid — left_index + 1); because the update query contains the whole range update not just a single value

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    9 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    I didn't understand, my mistake is here ?

    if(L != R)
{
   lazy[left(p)] += newValue;
   lazy[right(p)] += newValue;
}

    My code was based on this: http://www.geeksforgeeks.org/lazy-propagation-in-segment-tree/ Is it wrong too ?

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    • »
      »
      »
      9 years ago, # ^ |
      ← Rev. 4   Vote: I like it 0 Vote: I do not like it

      Solved it using the link. here's the AC code : http://ideone.com/kTHEpl Sorry I made a mistake. You can either Here's how I did it : http://ideone.com/nBUWgZ The only change in my code is that in my lazy update I'm updating the lazyval as the sum of subtree i.e : lazy[L] += ( lazy[stIndex]/(p.S — p.F + 1) )*(mid — p.F + 1); and icrementing the tree value by the lazy node i.e. tree[i] += lazy[i];

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        9 years ago, # ^ |
          Vote: I like it 0 Vote: I do not like it

        Thank's, my mistake was: Overflow, I changed my variables to long long and acc.

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    9 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Nope, as far as I understood lazy stores value to be added not sum of subtree.

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9 years ago, # |
  Vote: I like it 0 Vote: I do not like it

Here is my awful code.

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    9 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Thank you, but you know where is my mistake ?

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