线段树（更新区间查询区间）poj 3468 A Simple Problem with Integers

最新推荐文章于 2019-05-12 18:28:12 发布

Noble_le

最新推荐文章于 2019-05-12 18:28:12 发布

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分类专栏：线段树

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本文介绍了一个经典的区间操作问题，包括区间加法更新和区间求和查询，并提供了一段使用线段树实现的C++代码示例。

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题目链接：http://poj.org/problem?id=3468

A Simple Problem with Integers

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 119447		Accepted: 37100
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

代码：

#include<iostream>
#include<cstdio>
using namespace std;
typedef long long ll;
#define N 100000+10
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
ll sum[N<<2],add[N<<2];
struct Node
{
    int l,r;
    int mid()
    {
        return (l+r)>>1;
    }
}tree[N<<2];
void PushUp(int rt)
{
    sum[rt]=sum[rt<<1]+sum[rt<<1|1];
}

void build(int l,int r,int rt)
{
    tree[rt].l=l;
    tree[rt].r=r;
    add[rt]=0;
    if(l==r)
    {
        scanf("%I64d",&sum[rt]);
        return;
    }
    int m=tree[rt].mid();
    build(lson);
    build(rson);
    PushUp(rt);
}
void PushDown(int rt,int m)
{
    if(add[rt])
    {
        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;
    }
}
ll query(int l,int r,int rt)
{
    if(l==tree[rt].l&&r==tree[rt].r)
    {
        return sum[rt];
    }
    PushDown(rt,tree[rt].r-tree[rt].l+1);

    int m=tree[rt].mid();
    ll res=0;
    if(r<=m)
    {
        res+=query(l,r,rt<<1);
    }
    else if(l>m)
    {
        res+=query(l,r,rt<<1|1);
    }
    else
    {
        res+=query(l,m,rt<<1);
        res+=query(rson);
    }
    return res;
}
void update(int c,int l,int r,int rt)
{
    if(tree[rt].l==l&&tree[rt].r==r)
    {
        add[rt]+=c;
        sum[rt]+=(ll)c*(r-l+1);
        return ;
    }
   
    PushDown(rt,tree[rt].r-tree[rt].l+1);
    int m=tree[rt].mid();
    if(r<=m)
    {
        update(c,l,r,rt<<1);
    }
    else if(l>m)
    {
        update(c,l,r,rt<<1|1);
    }
    else
    {
        update(c,l,m,rt<<1);
        update(c,m+1,r,rt<<1|1);
    }
    PushUp(rt);
}
int main()
{
    int n,m;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        build(1,n,1);
        while(m--)
        {
            char ch[2];
            scanf("%s",ch);
            int a,b,c;
            if(ch[0]=='Q')
            {
                scanf("%d%d",&a,&b);
                printf("%lld\n",query(a,b,1));
            }
            else
            {
                scanf("%d%d%d",&a,&b,&c);
                update(c,a,b,1);
            }
        }
    }
    return 0;
}