POJ 1195- Mobile phones（二维BIT）

题意：

矩阵上的单点更新，范围求和

#include <map>
#include <set>
#include <list>
#include <cmath>
#include <queue>
#include <stack>
#include <cstdio>
#include <vector>
#include <string>
#include <cctype>
#include <complex>
#include <cassert>
#include <utility>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
using namespace std;
typedef pair<int,int> PII;
typedef long long ll;
#define lson l,m,rt<<1
#define pi acos(-1.0)
#define rson m+1,r,rt<<11
#define All 1,N,1
#define N 1100
#define read freopen("in.txt", "r", stdin)
const ll  INFll = 0x3f3f3f3f3f3f3f3fLL;
const int INF= 0x7ffffff;
const int mod =  1000000007;
int bit[N][N],n,m,a[N][N];
int lowbit(int x){
    return x&(-x);
}
void add(int x,int y,int d){
    for(int i=x;i<=n;i+=lowbit(i))
        for(int j=y;j<=n;j+=lowbit(j))
        bit[i][j]+=d;
}
int sum(int x,int y){
    int num=0;
    for(int i=x;i>0;i-=lowbit(i))
        for(int j=y;j>0;j-=lowbit(j))
        num+=bit[i][j];
    return num;
}
int main()
{
    int op,x1,x2,y1,y2,d;
    while(~scanf("%d",&op)){
        if(op==3)break;
        else if(op==0){
            scanf("%d",&n);
        memset(bit,0,sizeof(bit));
        memset(a,0,sizeof(a));
        }
        else if(op==1){
            scanf("%d%d%d",&x1,&y1,&d);
            x1++;
            y1++;
            if(a[x1][y1]+d<0)
                d=-a[x1][y1];
            add(x1,y1,d);
            a[x1][y1]+=d;
        }
        else{
            scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
            int tmp=sum(x2+1,y2+1)-sum(x2+1,y1)-sum(x1,y2+1)+sum(x1,y1);
            printf("%d\n",tmp);
        }
    }
return 0;
}