(简单) POJ 3468 A Simple Problem with Integers , 线段树+区间更新。

Description

  You have N integers, A1A2, ... , 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.

 

  题意很好理解,就是标准的线段树区间更新问题,可以说是模板。

  不过因为写的不熟练,被坑了两个WA,错的地方在代码里标出来了。。。

 

代码如下:

#include<iostream>
#include<cstdio>

#define lson L,M,po*2
#define rson M+1,R,po*2+1

using namespace std;

long long BIT[100005*4];
long long COL[100005*4];

void pushUP(int po)
{
    BIT[po]=BIT[po*2]+BIT[po*2+1];
}

void pushDown(int po,int len)
{
    if(COL[po])
    {
        BIT[po*2]+=(long long)COL[po]*(len-(len/2));
        BIT[po*2+1]+=(long long)COL[po]*(len/2);

        COL[po*2]+=COL[po];                 //注意是+=,不是= !!!
        COL[po*2+1]+=COL[po];
        COL[po]=0;
    }
}

void build_tree(int L,int R,int po)
{
    if(L==R)
    {
        COL[po]=0;
        cin>>BIT[po];
        return;
    }

    int M=(L+R)/2;

    build_tree(lson);
    build_tree(rson);

    pushUP(po);
}

long long query(int ql,int qr,int L,int R,int po)
{
    if(ql<=L&&qr>=R)
        return BIT[po];

    pushDown(po,(R-L+1));

    int M=(L+R)/2;

    if(qr<=M)
        return query(ql,qr,lson);
    if(ql>M)
        return query(ql,qr,rson);

    return query(ql,qr,rson)+query(ql,qr,lson);
}

void update(int ul,int ur,int add,int L,int R,int po)
{
    if(ul<=L&&ur>=R)
    {
        BIT[po]+=(long long)add*(R-L+1);
        COL[po]+=(long long)add;

        return;
    }

    pushDown(po,R-L+1);

    int M=(L+R)/2;

    if(ul<=M)
        update(ul,ur,add,lson);
    if(ur>M)
        update(ul,ur,add,rson);

    pushUP(po);
}

int main()
{
    int N,Q;
    char C;
    int a,b,c;

    while(~scanf("%d %d",&N,&Q))
    {
        build_tree(1,N,1);

        for(int i=0;i<Q;++i)
        {
            cin>>C;

            if(C=='Q')
            {
                scanf("%d %d",&a,&b);

                cout<<query(a,b,1,N,1)<<endl;
            }
            else
            {
                scanf("%d %d %d",&a,&b,&c);
                update(a,b,c,1,N,1);
            }
        }
    }

    return 0;
}
View Code

 

posted @ 2014-12-28 23:09  WhyWhy。  阅读(266)  评论(0编辑  收藏  举报