POJ 1195 Mobile Phones 二维树状数组

#include<iostream>
#include<math.h>
#include<string.h>
#include<queue>
#include<algorithm>
#include<cstdio>
#define rep(i,a,n) for(i=a;i<=n;i++)
#define per(i,a,n) for(i=n;i>=a;i--)
#define scd1(a) scanf("%d",&a)
#define scd2(a,b) scanf("%d%d",&a,&b)
#define ll long long
#define mem(a,b) memset(a,b,sizeof a);
using namespace std;

const int maxn=1024;
int T,cnt=0,N,M;
int mt[maxn+5][maxn+5];
int ist;

int lowbit(int num){
    return num&(-num);
}

void add(int x, int y, int d){
    while(x<=N){
        int j=y;
        while(j<=N){
            mt[x][j]+=d;j+=lowbit(j);
        }x+=lowbit(x);
    }
}

int sum(int x, int y){
    int ret=0;
    while(x>0){
        int j=y;
        while(j>0){
            ret+=mt[x][j];j-=lowbit(j);
        }
        x-=lowbit(x);
    }
    return ret;
}

int main(){
    while(~scd2(ist,N)){
        mem(mt,0);
        while(~scd1(ist)&&ist!=3){
            if(ist==1){
                int X,Y,A;
                scd2(X,Y);scd1(A);X++;Y++;
                add(X,Y,A);
            }
            if(ist==2){
                int L,B,R,T;
                scd2(L,B);scd2(R,T);L++;B++;R++;T++;
                int tmp=sum(R,T);
                tmp+=sum(L-1,B-1);
                tmp-=sum(L-1,T);
                tmp-=sum(R,B-1);
                printf("%d
",tmp);
            }
        }
    }
}

题意描述: 给出一个矩阵, 每次修改一个点的值, 每次质询要求回答一个子矩阵的数值和. 

还是裸的二维树状数组, 它的对偶题是POJ2155 Matrix 题解链接:http://www.cnblogs.com/LiXinze/p/7353436.html