POJ 3468 线段树

最新推荐文章于 2017-07-20 20:35:50 发布

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最新推荐文章于 2017-07-20 20:35:50 发布

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分类专栏：线段树

本文链接：https://blog.youkuaiyun.com/u011566113/article/details/9832973

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本文介绍了一种基于线段树的数据结构实现，用于处理数组区间内的批量加法操作及求和查询，通过懒惰标记优化更新效率。

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A Simple Problem with Integers

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

题意：C、将a-b的值全部加上c

Q：输出a-b的和

#include <iostream>
using namespace std;
const int maxn=255555;
long long sum[maxn<<2];
long long col[maxn<<2];
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
void pushup(int rt)
{
	sum[rt]=sum[rt<<1]+sum[rt<<1|1];
}
void pushdown(int rt,int m)
{
	if(col[rt])
	{
		col[rt<<1]+=col[rt];//一开始脸都写成等于，wa掉了
		col[rt<<1|1]+=col[rt];
		sum[rt<<1]+=(m-(m>>1))*col[rt];
		sum[rt<<1|1]+=(m>>1)*col[rt];
		col[rt]=0;
	}
}
void build(int l,int r,int rt)
{
	col[rt]=0;
	if(l==r)
	{
		scanf("%lld",&sum[rt]);
		return;
	}
	int m=(l+r)>>1;
	build(lson);
	build(rson);
	pushup(rt);
}
void update(int l,int r,int rt,int L,int R,int w)
{
	if(L<=l&&R>=r)
	{
		col[rt]+=w;
		sum[rt]+=w*(r-l+1);
		return;
	}
	pushdown(rt,r-l+1);
	int m=(l+r)>>1;
	if(L<=m)update(lson,L,R,w);
	if(R>m)update(rson,L,R,w);
	pushup(rt);
}
long long query(int l,int r,int rt,int L,int R)
{
	if(L<=l&&R>=r)
		return sum[rt];
	pushdown(rt,r-l+1);//遗漏了这个 好无语
	int m=(l+r)>>1;
	long long ret=0;
	if(L<=m)ret+=query(lson,L,R);
	if(R>m)ret+=query(rson,L,R);
	return ret;
}
int main()
{
	int n,m,a,b,cc;
	char c[2];
	while(~scanf("%d%d",&n,&m))
	{
		build(1,n,1);
		while(m--)
		{
			scanf("%s",c);
			if(c[0]=='Q')
			{
				scanf("%d%d",&a,&b);
				printf("%lld\n",query(1,n,1,a,b));
			}
			else
			{
				scanf("%d%d%d",&a,&b,&cc);
				update(1,n,1,a,b,cc);
			}
		}
	}
	return 0;
}