POJ 3468 线段树插线问线区间求和

最新推荐文章于 2012-08-01 17:40:55 发布

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最新推荐文章于 2012-08-01 17:40:55 发布

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原文链接：http://www.cnblogs.com/javaspring/archive/2012/04/20/2656369.html

本文详细解析了线段树算法的应用场景及实现细节，通过一个具体的编程问题——ASimpleProblemwithIntegers，介绍了如何利用线段树进行区间更新和查询操作。

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典型的线段树题目，插线问线，求区间和。注意中间会超int范围

题目：

A Simple Problem with Integers

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 29416		Accepted: 8241
Case 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 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

ac代码：

#include <iostream>
#include <cstdio>
using namespace std;
const int N=100010;
struct tree{
	int left,right;
	long long add,sum;
}tt[N*4];
int num[N];
void built_tree(int lp,int rp,int pos){
	tt[pos].left=lp;
	tt[pos].right=rp;
	tt[pos].add=0;
	if(lp==rp){
	  tt[pos].sum=num[lp];
	  return;
	}
	int mid=(tt[pos].left+tt[pos].right)/2;
	built_tree(lp,mid,pos*2);
	built_tree(mid+1,rp,pos*2+1);
	tt[pos].sum=tt[pos*2].sum+tt[pos*2+1].sum;
}
void update(int lp,int rp,int add,int pos){
	if(lp<=tt[pos].left&&rp>=tt[pos].right){
		tt[pos].add+=add;
		tt[pos].sum+=(tt[pos].right-tt[pos].left+1)*add;
		return;
	}
	if(tt[pos].add){
		tt[pos*2].sum+=(tt[pos*2].right-tt[pos*2].left+1)*tt[pos].add;
		tt[pos*2+1].sum+=(tt[pos*2+1].right-tt[pos*2+1].left+1)*tt[pos].add;
		tt[pos*2].add+=tt[pos].add;
		tt[pos*2+1].add+=tt[pos].add;
		tt[pos].add=0;
	}
	int mid=(tt[pos].right+tt[pos].left)/2;
	if(lp<=mid)
		update(lp,rp,add,pos*2);
	if(rp>mid)
		update(lp,rp,add,pos*2+1);
	tt[pos].sum=tt[pos*2].sum+tt[pos*2+1].sum;
}
long long find(int lp,int rp,int pos){
	if(lp==tt[pos].left&&rp==tt[pos].right)
		return tt[pos].sum;
	if(tt[pos].add){
	  	tt[pos*2].sum+=(tt[pos*2].right-tt[pos*2].left+1)*tt[pos].add;
		tt[pos*2+1].sum+=(tt[pos*2+1].right-tt[pos*2+1].left+1)*tt[pos].add;
		tt[pos*2].add+=tt[pos].add;
		tt[pos*2+1].add+=tt[pos].add;
		tt[pos].add=0;
	}
	int mid=(tt[pos].left+tt[pos].right)/2;
	if(lp>mid)
		return find(lp,rp,pos*2+1);
	else if(rp<=mid)
		return find(lp,rp,pos*2);
	else
		return find(lp,mid,pos*2)+find(mid+1,rp,pos*2+1);
}
int main(){
	int n,m;
	while(~scanf("%d%d",&n,&m)){
	  for(int i=1;i<=n;++i)
		  scanf("%d",&num[i]);
	  built_tree(1,n,1);
	  char ss[2];
	  int x,y,z;
	  while(m--){
	    scanf("%s",ss);
		if(ss[0]=='C'){
		  scanf("%d%d%d",&x,&y,&z);
		  update(x,y,z,1);
		}
		else{
		  scanf("%d%d",&x,&y);
		  long long total=find(x,y,1);
		  printf("%lld\n",total);
		}
	  }
	}
	return 0;
}

转载于:https://www.cnblogs.com/javaspring/archive/2012/04/20/2656369.html