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POSegment tree Lazy propagation & my code
 

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  1. POSegment tree Lazy propagation & my code
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    <p>I am currently solving a problem on segment tree. I think the problem needs lazy propagation concept to be solved. As I'm very new to this concept, i'm having trouble with my code.</p> <p>The problem in a nutshell is as follows:</p> <p>initially, all array elements are 0 and they are indexed 0 to N-1</p> <p>command 1. 0 x y v - updates value of each array indexes between x and y by v</p> <p>command 2. 1 x y - output the sum of all numbers between array index x &amp; y. </p> <p>Input starts with an integer T (≤ 5), denoting the number of test cases.</p> <p>Each case contains two integers n (1 ≤ n ≤ 105) and q (1 ≤ q ≤ 50000). Each of the next q lines contains a task in one of the following form:</p> <p>0 x y v (0 ≤ x ≤ y &lt; n, 1 ≤ v ≤ 1000)</p> <p>1 x y (0 ≤ x ≤ y &lt; n) For each case, print the case number first. Then for each query '1 x y', print the sum of all the array elements between x and y. Here is my attempt:</p> <pre class="lang-c prettyprint-override"><code>template&lt;class T&gt; class SegmentTree { T *tree,*update_tree; long size; public: SegmentTree(long N) { long x= (long)ceil(log2(N))+1; long size = 2*(long)pow(2,x); tree = new T[size]; update_tree = new T[size]; memset(tree,0,sizeof(tree)); memset(update_tree,0,sizeof(update_tree)); } void update(long node, long start, long end, long i, long j, long val) { if(start&gt;j || end&lt;i) return; if(start&gt;=i &amp;&amp; end&lt;=j){ if(start==end){ tree[node]+=val; return; } tree[node]+=val; update_tree[2*node] += val; update_tree[2*node+1]+=val; return; } long mid = (start+end)/2; update(2*node,start,mid,i,j,val); update(2*node+1,mid+1,end,i,j,val); } T query(long node, long start, long end, long i, long j, long val) { if(start&gt;j || end&lt;i) return -1; if(start&gt;=i &amp;&amp; end&lt;=j) return ((tree[node]+val)*(end-start+1)); long a,b; a = update_tree[2*node]; b = update_tree[2*node+1]; long mid = (start+end)/2; long val1 = query(2*node,start,mid,i,j,val+a); long val2 = query(2*node+1,mid+1,end,i,j,val+b); if(val1==-1) return val2; if(val2==-1) return val1; return val1+val2; } }; int main() { long N,q,x,y,res; int tc=1, T,v,d; scanf("%d",&amp;T); while(tc&lt;=T) { scanf("%ld %ld",&amp;N,&amp;q); SegmentTree&lt;long&gt;s(N); printf("Case %d:\n",tc++); while(q--){ scanf("%d",&amp;d); if(!d){ scanf("%ld %ld %d",&amp;x,&amp;y,&amp;v); s.update(1,0,N-1,x,y,v); } else{ scanf("%ld %ld",&amp;x,&amp;y); res = s.query(1,0,N-1,x,y,0); printf("%ld\n",res); } } } return 0; } </code></pre>
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    1. USuser1871166
    1. USuser1941259
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      1. POSegment tree Lazy propagation & my code
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      1. VTUpMod
    2. VO
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      1. POSegment tree Lazy propagation & my code
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    1. CODo you want lazy propagation (create elements only when they're going to be accessed/modified) or lazy evaluation (modify elements only when you try to read their value)? It seems your trying to achieve the latter.
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      1. POSegment tree Lazy propagation & my code
      1. USZeta
    2. COI don't have any idea on lazy evaluation, just heard the term propagation and tried to code it. I need to solve the problem, which has less complexity.
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      1. POSegment tree Lazy propagation & my code
      1. USuser1941259
    3. COPerhaps it is overflowing? The worst case query could require an answer of 105*50000*1000 which is more than will fit in a 32bit long.
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      1. POSegment tree Lazy propagation & my code
      1. USPeter de Rivaz
 

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