codeforces 390E Inna and Large Sweet Matrix

本题的主要算法就是区间更新和区间求和;

可以用线段树和树状数组来做;

感觉线段树写的太麻烦了,看到官方题解上说可以用树状数组做,觉得很神奇,以前用过的树状数组都是单点维护,区间求和的;

其实树状数组还可以区间维护,单点求值;和区间维护,区间求和的;

详情请见博客

#include<cstdio>
#include<cstring>
#include<cmath>
#include<iostream>
#define maxn 4000010
#define ll long long
using namespace std;

ll a[2][maxn];
ll b[2][maxn];

void add_a(int flag,int x,ll value)
{
    while(x>0)
    {
        a[flag][x]+=value;
        x-=x&(-x);
    }
}

void add_b(int flag,int n,int x,ll value)
{
    for(int i=x;i<=n;i+=i&(-i))
        b[flag][i]+=x*value;
}

ll sum_a(int flag,int n,int x)
{
    ll sum=0;
    while(x<=n)
    {
        sum+=a[flag][x];
        x+=x&(-x);
    }
    return sum;
}

ll sum_b(int flag,int x)
{
    ll sum=0;
    while(x>0)
    {
        sum+=b[flag][x];
        x-=x&(-x);
    }
    return sum;
}

ll getsum(int flag,int n,int x)
{
    if(x)
        return sum_a(flag,n,x)*x+sum_b(flag,x-1);
    else
    {
        return 0;
    }
}

int main()
{
    int n,m,w;
    int comand;
    int x1,x2,y1,y2;
    int v;
    scanf("%d%d%d",&n,&m,&w);
    for(int i=0; i<w; i++)
    {
        scanf("%d",&comand);
        if(comand==0)
        {
            scanf("%d%d%d%d%d",&x1,&y1,&x2,&y2,&v);
            ll vy=v*(y2-y1+1);
            ll vx=v*(x2-x1+1);
            add_a(0,x2,vy);
            add_b(0,n,x2,vy);
            if(x1>1)
            {
                add_a(0,x1-1,-vy);
                add_b(0,n,x1-1,-vy);
            }
            add_a(1,y2,vx);
            add_b(1,m,y2,vx);
            if(y1>1)
            {
                add_a(1,y1-1,-vx);
                add_b(1,m,y1-1,-vx);
            }
        }
        else
        {
            scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
            ll tmp1=getsum(0,n,x2)-getsum(0,n,x1-1);
            ll tmp2=getsum(1,m,y1-1);
            ll tmp3=getsum(1,m,m)-getsum(1,m,y2);
            cout<<tmp1-tmp2-tmp3<<endl;
        }
    }
    return 0;
}
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posted @ 2014-02-23 21:29  Yours1103  阅读(200)  评论(0编辑  收藏  举报