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Need help in segment tree problem
By acash, history, 7 years ago, In English

I tried many test cases ,I am passing all of them But it is failing at test case 5 on codeforces. Since test case are quite large ,I am unable to figure out my error.Please forgive me if my question is not good. 55748387(My submission)

https://codeforces.me/contest/52/problem/C

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7 years ago, hide # |
 
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Change removelazy(node1, l, r) to removelazy(node1, start, end1)

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    7 years ago, hide # ^ |
     
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    Still getting wrong ,should i check also if the node is lazy or not while answering the query? Because i am not checking

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7 years ago, hide # |
 
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I added the removelazy in query part

I think there are many other issues in the implementation. I suggest you to check your implementation once again.

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7 years ago, hide # |
 
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https://codeforces.me/contest/52/submission/54744859 take a look at my submission

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    7 years ago, hide # ^ |
     
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    Please can you tell me where I am doing the mistake?

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      7 years ago, hide # ^ |
       
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      First of all you are consider 0 base indexing in your build() function so your left child is (node1 << 1) + 1 and right child is (node1 << 1) + 2 and your query() function is

      node query(int node1, int l, int r, int start, int end1) {

      if (tree[node1].lazy) {
    removelazy(node1, l, r);
}

node req;
if (l > end1 || r < start)return req;
if (l <= start && end1 <= r)return tree[node1];
int mid = (start + end1) >> 1;
node a = query((node1 << 1), l, r, start, mid);
node b = query((node1 << 1) + 1, l, r, mid + 1, end1);
merge1(req, a, b);
return req;

      }

      You do not consider if(tree[node1].lazy != 0) case in your query function

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