Time Limit: 5000MS | Memory Limit: 131072K | |
Total Submissions: 102132 | Accepted: 31872 | |
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. ⑴000000000 ≤ 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. ⑴0000 ≤ 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
原题链接:http://poj.org/problem?id=3468
线段树区间更新入门题。区间更新用单点更新的方法去做必定会TLE,当要更新的区间正好是某个节点的区间的时候,我们就更新到此,不再继续往下更新,从而节省了时间,实现了次区间更新的高效性。
但后面更新的区间是上次更新区间的子区间的时候就要把之前保存在父节点的更新数据“下放”到对应的子区间,从而实现了操作的正确性。
对线段树区间更新的描写网上有好多不错的博客,我就不用再去造轮子了。写法也有好多种,合适自己的才是最好的。
以下3篇代表了几种不同的写法。
此题参考博客:http://blog.csdn.net/w00w12l/article/details/7920846
http://blog.csdn.net/acdreamers/article/details/8578161
http://blog.csdn.net/acceptedxukai/article/details/6933446
#include <iostream> #include <cstdio> using namespace std; const int maxn=100000+5; #define lson l,mid,rt<<1 #define rson mid+1,r,rt<<1|1 typedef long long LL; LL sum[maxn<<2],add[maxn]; int a[maxn]; struct node { int l,r,m; LL sum,mark; }tree[maxn<<2]; void BuildTree(int l,int r,int k) { tree[k].l=l; tree[k].r=r; tree[k].m=(l+r)>>1; tree[k].mark=0; if(l==r) { tree[k].sum=a[l]; return; } int mid=(l+r)>>1; BuildTree(l,mid,k<<1); BuildTree(mid+1,r,k<<1|1); tree[k].sum=(tree[k<<1].sum+tree[k<<1|1].sum); } void UpdateTree(int l,int r,int k,int x) { if(tree[k].l==l&&tree[k].r==r) { tree[k].mark+=x; return; } tree[k].sum+=(LL)x*(r-l+1); //int mid=(tree[k].l+tree[k].r)>>1; int mid=tree[k].m; if(r<=mid) UpdateTree(l,r,k<<1,x); else if(mid<l) UpdateTree(l,r,k<<1|1,x); else { UpdateTree(l,mid,k<<1,x); UpdateTree(mid+1,r,k<<1|1,x); } } LL QueryTree(int l,int r,int k) { //cout<<l<<","<<r<<","<<k<<endl; if(tree[k].l==l&&tree[k].r==r) return tree[k].sum+tree[k].mark*(LL)(r-l+1); if(tree[k].mark!=0) { tree[k<<1].mark+=tree[k].mark; tree[k<<1|1].mark+=tree[k].mark; tree[k].sum+=(LL)(tree[k].r-tree[k].l+1)*tree[k].mark;//记得+1 tree[k].mark=0; } //int mid=(tree[k].l+tree[k].r)>>1; int mid=tree[k].m; if(mid>=r) return QueryTree(l,r,k<<1); else if(l>mid) return QueryTree(l,r,k<<1|1); else return QueryTree(l,mid,k<<1)+QueryTree(mid+1,r,k<<1|1); } int main() { int n,m; while(cin>>n>>m) { for(int i=1;i<=n;i++) cin>>a[i]; BuildTree(1,n,1); char ch[2]; int x,y,z; while(m--) { scanf("%s",ch); if(ch[0]=='Q') { scanf("%d%d",&x,&y); printf("%lld\n",QueryTree(x,y,1)); } else if(ch[0]=='C') { scanf("%d%d%d",&x,&y,&z); UpdateTree(x,y,1,z); } } } return 0; }