poj 3468 A Simple Problem with Integers(线段树+区间更新+区间求和)
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题目链接:http://poj.org/problem?id=3468
Time Limit: 5000MS | Memory Limit: 131072K | |
Total Submissions: 83959 | Accepted: 25989 | |
Case Time Limit: 2000MS |
Description
You have N integers, A1, A2, ... , AN. You need to deal with two kinds of operations. One type of operation is to add some given number to each number in a given interval. The other is to ask for the sum of numbers in a given interval.
Input
The first line contains two numbers N and Q. 1 ≤ N,Q ≤ 100000.
The second line contains N numbers, the initial values of A1, A2, ... , AN. -1000000000 ≤ Ai ≤ 1000000000.
Each of the next Q lines represents an operation.
"C a b c" means adding c to each of Aa, Aa+1, ... , Ab. -10000 ≤ c ≤ 10000.
"Q a b" means querying the sum of Aa, Aa+1, ... , Ab.
Output
You need to answer all Q commands in order. One answer in a line.
Sample Input
10 5 1 2 3 4 5 6 7 8 9 10 Q 4 4 Q 1 10 Q 2 4 C 3 6 3 Q 2 4
Sample Output
4 55 9 15
Hint
Source
对于更新树是为了避免改动到最底下而导致超时问题。所以每次改动仅仅改动相相应的区间就可以。然后记录一个add。下次更新或者查询的时候,假设查到该节点,就把add直接加到子节点上去,在将add变为0,避免下次还会反复加。这样仅仅更新到查询的子区间,不须要再往下找了,所以时间复杂度为O(n),更新树和查询树都须要这样。
由于add不为0,该add从根一直加到了该节点,之前的都加过了,假设更新到时候不加到子节点。还要通过子节点更新当前节点,当前节点的sum值里面含有的add就会被“抹掉”,就不能保证正确性了。还须要注意的就是要用__int64。
#include <iostream> #include <cstdio> using namespace std; #define LL __int64 struct node { int l,r; LL sum; LL add; //int flag;//用来表示有几个加数 } s[100000*4]; void InitTree(int l,int r,int k) { s[k].l=l; s[k].r=r; s[k].sum=0; s[k].add=0; if (l==r) return ; int mid=(l+r)/2; InitTree(l,mid,2*k); InitTree(mid+1,r,2*k+1); } void UpdataTree(int l,int r,LL add,int k) { if (s[k].l==l&&s[k].r==r) { s[k].add+=add; s[k].sum+=add*(r-l+1); return ; } if (s[k].add!=0)//加数为0就不须要改变了 { s[2*k].add+=s[k].add; s[2*k+1].add+=s[k].add; s[2*k].sum+=s[k].add*(s[2*k].r-s[2*k].l+1); s[2*k+1].sum+=s[k].add*(s[2*k+1].r-s[2*k+1].l+1); s[k].add=0; } int mid=(s[k].l+s[k].r)/2; if (l>mid) UpdataTree(l,r,add,2*k+1); else if (r<=mid) UpdataTree(l,r,add,2*k); else { UpdataTree(l,mid,add,2*k); UpdataTree(mid+1,r,add,2*k+1); } s[k].sum=s[2*k].sum+s[2*k+1].sum; } LL SearchTree(int l,int r,int k) { if (s[k].l==l&&s[k].r==r) return s[k].sum; if (s[k].add!=0) { s[2*k].add+=s[k].add; s[2*k+1].add+=s[k].add; s[2*k].sum+=s[k].add*(s[2*k].r-s[2*k].l+1); s[2*k+1].sum+=s[k].add*(s[2*k+1].r-s[2*k+1].l+1); s[k].add=0; } int mid=(s[k].l+s[k].r)/2; if (l>mid) return SearchTree(l,r,2*k+1); else if (r<=mid) return SearchTree(l,r,2*k); else return SearchTree(l,mid,2*k)+SearchTree(mid+1,r,2*k+1); } int main() { int n,q; LL w; while (~scanf("%d%d",&n,&q)) { InitTree(1,n,1); for (int i=1; i<=n; i++) { scanf("%lld",&w); UpdataTree(i,i,w,1); } for (int i=1; i<=q; i++) { char ch; int a,b; LL c; getchar(); scanf("%c%d%d",&ch,&a,&b); if (ch=='C') { scanf("%lld",&c); UpdataTree(a,b,c,1); } else if (ch=='Q') { LL ans=SearchTree(a,b,1); printf ("%lld\n",ans); } } } return 0; }
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