SPOJ1029 - Matrix Summation

题目大意

给定一个N*N的矩阵A，每个元素的初始值为0，可以对矩阵进行一下两种操作：

1、修改A[i][j]的值为d,(1<=i,j<=N)

2、查询左下角坐标为(x1,y1)，右上角坐标为(x2,y2)的子矩阵的元素和

题解

基本的树状数组，单点增减，区间求和

#include<iostream>
#include<cstring>
#include<cstdio>
#define MAXN 1050
using namespace std;
int n;
int c[MAXN][MAXN],a[MAXN][MAXN];
int lowbit(int x)
{
    return x&-x;
}
void add(int x,int y,int d)
{
    int i,j;
    for(i=x;i<=MAXN;i+=lowbit(i))
    for(j=y;j<=MAXN;j+=lowbit(j))
    c[i][j]+=d;
}
int sum(int x,int y)
{
    int ret=0,i,j;
    for(i=x;i>0;i-=lowbit(i))
    for(j=y;j>0;j-=lowbit(j))
        ret+=c[i][j];
    return ret;
}
int main()
{
    int T,x1,x2,y1,y2,ans,t;
    char s[5];
    scanf("%d",&T);
    while(T--)
    {
        memset(c,0,sizeof(c));
        memset(a,0,sizeof(a));
         scanf("%d",&n);
         while(scanf("%s",s)!=EOF)
         {
             if(s[0]=='E') break;
             if(s[2]=='T')
             {
                 scanf("%d%d%d",&x1,&y1,&t);
                 x1++;
                 y1++;
                 add(x1,y1,t-a[x1][y1]);
                 a[x1][y1]=t;
             }
             else
             {
                 scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
                 x1++;
                 y1++;
                 x2++;
                 y2++;
                 ans=sum(x2,y2)-sum(x1-1,y2)-sum(x2,y1-1)+sum(x1-1,y1-1);
                 printf("%d\n",ans);
             }
         }
    }
    return 0;
}