POJ3468 A Simple Problem with Integers(线段树成段更新,区间查询)

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A Simple Problem with Integers

Time Limit: 5000MS Memory Limit: 131072KTotal Submissions: 72201 Accepted: 22282Case Time Limit: 2000MS

Description

You have N integers, A₁, A₂, ... , A_N. 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, ... , A_N. -1000000000 ≤ A_i ≤ 1000000000.
Each of the next Q lines represents an operation.
"C a b c" means adding c to each of A_a, A_a₊₁, ... , A_b. -10000 ≤ c ≤ 10000.
"Q a b" means querying the sum of A_a, A_a₊₁, ... , A_b.

Output

You need to answer all Q commands in order. One answer in a line.

Sample Input

10 51 2 3 4 5 6 7 8 9 10Q 4 4Q 1 10Q 2 4C 3 6 3Q 2 4

Sample Output

解题思路:

本题一看，AC数那么高，其实感觉手写还是不易，关键理解线段树成段更新，区间查询的要点就行，很多人估计是套模版过的吧，但是还是理解一下线段树的一些策略比较好，比如懒惰处理，为什么需要懒惰处理(pushdown)呢，因为如果每层都更新的话太花费时间了，因为有时你并不需要查询所有的结点，所以要查询的时候再去更新结点就行，所以就有了懒惰处理。其他的倒是没有什么值得注意的地方，套线段树的模版就行。

AC代码:

#include<iostream>using namespace std;const int maxn = 100010;typedef long long LL;#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1LL add[maxn<<2]; //4 * maxnLL sum[maxn<<2];void PushUP(int rt){sum[rt] = sum[rt<<1] + sum[rt<<1|1];}void PushDown(int rt,int m)  //懒惰，往下传递值，子节点需要查询的时候再更新 {if(add[rt]){add[rt<<1] += add[rt];  //接受父节点传递来的值 add[rt<<1|1] += add[rt]; sum[rt<<1] += (m>>1) * add[rt]; //父节点有2*N个子节点，那么子节点就要/2为N sum[rt<<1|1] += (m>>1) * add[rt];add[rt] = 0; //父节点的值传递后，要清空 }}void Build(int l,int r,int rt){add[rt] = 0;if(l == r){cin>>sum[rt];return;}int m = (l + r) / 2;Build(lson);Build(rson);PushUP(rt);  //等价于父节点等于两个子节点的和 }void Update(int L,int R,int c,int l,int r,int rt)  //c为增加的值 {if(L <= l && R >= r){add[rt] += c;sum[rt] += (LL)c * (r - l + 1); //每个数+c,区间中一共有r - l + 1个return; }PushDown(rt,r-l+1);//加上后懒惰传递.int m = (l + r) >> 1;if(L <= m)Update(L,R,c,lson); //更新左子树if(R > m)Update(L,R,c,rson); //更新右子树PushUP(rt); //等价于父节点等于两个子节点的和 }LL Query(int L,int R,int l,int r,int rt) //L,R为需要查询的区间 {if(L <= l && R >= r){return sum[rt];}PushDown(rt,r-l+1); //当前结点和区间内一共有多少个结点, int m = (l + r) >> 1; LL ret = 0;if(L <= m) ret += Query(L,R,lson);if(R > m) ret += Query(L,R,rson);return ret;}int main(){int m,n;int i;char str;while(cin>>m>>n){Build(1,m,1);while(n--){int num1,num2,num3;cin>>str;if(str == 'Q'){cin>>num1>>num2;cout<<Query(num1,num2,1,m,1)<<endl;}else{cin>>num1>>num2>>num3;Update(num1,num2,num3,1,m,1);}}}return 0;}

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