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| A Simple Problem with Integers | ||||||
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| 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 | ||||||
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Input contain multiple test cases,for each case: The first line contains two numbers N and Q. 1 ≤ N,Q ≤ 100000. |
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| Output | ||||||
| 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. | ||||||
| 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 |
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| Sample Output | ||||||
4 55 9 15 |
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| Source | ||||||
| POJ Monthly--2007.11.25, Yang Yi | ||||||
| Recommend | ||||||
| `Wind @Hrbust |
线段树区间更新，如果真的全部更新每个节点，会很耗费时间，所以在可行的节点上标记一下，这样下次查询的时候如果查询到这个节点，就往下更新一次
#include<cstdio> #include<cmath> #include<queue> #include<iostream> using namespace std; const int m=100000; typedef long long LL; struct node { LL Inc; LL sum; }a[m<<2]; LL b[m+1]; void build(int L,int R,int rt) { a[rt].Inc=0; if(L==R) { // scanf("%lld",&b[L]); a[rt].sum=b[L]; return ; } int lr=rt<<1,rr=rt<<1|1; int mid=(L+R)>>1; build(L,mid,lr); build(mid+1,R,rr); a[rt].sum=a[lr].sum+a[rr].sum; } void update(int l,int r,int L,int R,int rt,LL x) { if(L==l&&R==r) { a[rt].Inc+=x; return ; } a[rt].sum+=(r-l+1)*x; int m=(L+R)>>1; int lr=rt<<1,rr=rt<<1|1; if(r<=m) { update(l,r,L,m,lr,x); } else if(l>m) { update(l,r,m+1,R,rr,x); } else { update(l,m,L,m,lr,x); update(m+1,r,m+1,R,rr,x); } } LL query(int l,int r,int L,int R,int rt) { if(l<=L&&r>=R) { return a[rt].sum+a[rt].Inc*(r-l+1); } a[rt].sum+=(R-L+1)*a[rt].Inc; int m=(L+R)>>1; int lr=rt<<1,rr=rt<<1|1; if(a[rt].Inc) { //update(L,m,L,m,lr,a[rt].Inc); // update(m+1,R,m+1,R,rr,a[rt].Inc); a[lr].Inc+=a[rt].Inc; a[rr].Inc+=a[rt].Inc; } a[rt].Inc=0; if(r<=m) { return query(l,r,L,m,lr); } else if(l>m) { return query(l,r,m+1,R,rr); } LL left=query(l,m,L,m,lr); LL right=query(m+1,r,m+1,R,rr); return left+right; } int main() { int N,Q; while(~scanf("%d%d",&N,&Q)) { int i; for(i=1;i<=N;i++) { scanf("%lld",&b[i]); } build(1,N,1); char c; int l,r; LL x; while(Q--) { cin>>c; if(c==‘Q‘) { scanf("%d%d",&l,&r); printf("%lld\n",query(l,r,1,N,1)); } else if(c==‘C‘) { // cout<<"dsfdsf"<<endl; scanf("%d%d%lld",&l,&r,&x); update(l,r,1,N,1,x); } } } return 0; }
A Simple Problem with Integers(区间更新)
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原文地址：http://www.cnblogs.com/beige1315402725/p/5001632.html
