[ACM] poj 3468 A Simple Problem with Integers（段树，为段更新，懒惰的标志）

区间加查询问题

最新推荐文章于 2019-07-01 16:43:05 发布

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最新推荐文章于 2019-07-01 16:43:05 发布

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原文链接：http://www.cnblogs.com/zfyouxi/p/4864869.html

本文介绍了一个涉及区间操作的问题，包括区间内元素统一增加特定数值及求区间和的操作，并提供了一种有效的解决方案。通过使用线段树结构进行懒惰传播，实现了高效地处理大量更新和查询请求。

A Simple Problem with Integers

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 55273		Accepted: 16628
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

Hint

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

Source

POJ Monthly--2007.11.25, Yang Yi

解题思路：

某个区间内的数都加上同一个数，成段更新。

代码：

#include <iostream>
#include <stdio.h>
#include <algorithm>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
const int maxn=100005;
long long sum[maxn<<2];
long long add[maxn<<2];
int n,k;
void PushUp(int rt)
{
    sum[rt]=sum[rt<<1]+sum[rt<<1|1];
}
void PushDown(int rt,int m)
{
    if(add[rt])
    {
        //add[rt<<1]=add[rt<<1|1]=add[rt];//不能写成这样
        add[rt<<1]+=add[rt];
        add[rt<<1|1]+=add[rt];
        sum[rt<<1]+=add[rt]*(m - (m>>1));
        sum[rt<<1|1]+=add[rt]*(m>>1);
        add[rt]=0;
    }
}
void build(int l,int r,int rt)
{
    add[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 c,int l,int r,int rt)
{
    if(L<=l&&r<=R)
    {
        add[rt]+=c;
        sum[rt]+=(long long)c*(r-l+1);
        return;
    }
    PushDown(rt,r-l+1);
    int m=(r+l)>>1;
    if(L<=m)
        update(L,R,c,lson);
    if(R>m)
        update(L,R,c,rson);
    PushUp(rt);
}
long long query(int L,int R,int l,int r,int rt)//注意long long类型
{
    if(L<=l&&r<=R)
    {
        return sum[rt];
    }
    PushDown(rt,r-l+1);//查询的时候要向下更新。懒惰标记
    int m=(l+r)>>1;
    long long  ans=0;//注意long long
    if(L<=m)
        ans+=query(L,R,lson);
    if(R>m)
        ans+=query(L,R,rson);
    return ans;
}
int main()
{
    scanf("%d%d",&n,&k);
    build(1,n,1);
    char c;int a,b,d;
    while(k--)
    {
        scanf("%s",&c);
        if(c=='Q')
        {
            scanf("%d%d",&a,&b);
            printf("%lld\n",query(a,b,1,n,1));
        }
        else
        {
            scanf("%d%d%d",&a,&b,&d);
            update(a,b,d,1,n,1);
        }
    }
    return 0;
}

转载于:https://www.cnblogs.com/zfyouxi/p/4864869.html