POJ 3468 线段树

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我是靠谱客的博主现代樱桃，最近开发中收集的这篇文章主要介绍POJ 3468 线段树，觉得挺不错的，现在分享给大家，希望可以做个参考。

概述

http://poj.org/problem?id=3468

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

Hint

The sums may exceed the range of 32-bit integers.

题目大意：给出n个数和m个操作，Q a b，则输出区间[a，b]的和；Q a b c，则把区间[a，b]的所有元素都加上c。

思路：线段树，因为涉及到区间修改，所以要用到lazy标记。

线段树基本知识：https://blog.csdn.net/xiji333/article/details/87973714

#include<iostream>
#include<cstdio>
#define INF 0x3f3f3f3f
typedef long long ll;
using namespace std;
struct node
{
ll l,r;
ll sum;
ll lazy;	//lazy标记
};
node tree[100000*4+5];
ll a[100005];
ll n,m;
void build(ll i,ll l,ll r)
{
tree[i].l=l,tree[i].r=r;
if(l==r)
{
tree[i].sum=a[l];
return ;
}
ll mid=(l+r)>>1;
build(i<<1,l,mid);	//左子树
build(i<<1|1,mid+1,r);	//右子树
tree[i].sum=tree[i<<1].sum+tree[i<<1|1].sum;//区间和
}
void down(ll i)//节点i的标记下传
{
tree[i<<1].lazy+=tree[i].lazy;	//lazy标记下传
tree[i<<1|1].lazy+=tree[i].lazy;
tree[i<<1].sum+=tree[i].lazy*(tree[i<<1].r-tree[i<<1].l+1);//修改值
tree[i<<1|1].sum+=tree[i].lazy*(tree[i<<1|1].r-tree[i<<1|1].l+1);
tree[i].lazy=0;
}
void update(ll i,ll l,ll r,ll v)
{
if(tree[i].l==l&&tree[i].r==r)//要修改的区间就是当前区间
{
tree[i].sum+=(tree[i].r-tree[i].l+1)*v;//修改区间值
tree[i].lazy+=v;	//修改lazy标记
return ;
}
if(tree[i].lazy)//走到这一步说明要用到子节点了 lazy下传
down(i);
ll mid=(tree[i].l+tree[i].r)>>1;
if(r<=mid)//左半区间
update(i<<1,l,r,v);
else if(l>=mid+1)//右半区间
update(i<<1|1,l,r,v);
else //左右均有
{
update(i<<1,l,mid,v);
update(i<<1|1,mid+1,r,v);
}
tree[i].sum=tree[i<<1].sum+tree[i<<1|1].sum;
}
ll query(ll i,ll l,ll r)
{
if(tree[i].l==l&&tree[i].r==r)//要查询的区间就是当前区间
return tree[i].sum;
if(tree[i].lazy)//要用到子节点 lazy下传
down(i);
ll mid=(tree[i].l+tree[i].r)>>1;
if(r<=mid)//左半部分
return query(i<<1,l,r);
else if(l>=mid+1)	//右半部分
return query(i<<1|1,l,r);
else //左右均有
return query(i<<1,l,mid)+query(i<<1|1,mid+1,r);
}
int main()
{
scanf("%lld %lld",&n,&m);
for(ll i=1;i<=n;i++)
scanf("%lld",&a[i]);
build(1,1,n);
char ch;
ll t1,t2,t3;
for(ll i=0;i<m;i++)
{
getchar();
ch=getchar();
if(ch=='Q')
{
scanf("%lld%lld",&t1,&t2);
printf("%lldn",query(1,t1,t2));
}
else if(ch=='C')
{
scanf("%lld%lld%lld",&t1,&t2,&t3);
update(1,t1,t2,t3);
}
}
return 0;
}