POJ 2155 Matrix(二维树状数组)

Matrix Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 28817   Accepted: 10507 Description Given an N*N matrix A, whose elements are either 0 or 1. A[i, j] means the number in the i-th row and j-th column. Initially we have A[i, j] = 0 (1 <= i, j <= N).  We can change the matrix in the following way. Given a rectangle whose upper-left corner is (x1, y1) and lower-right corner is (x2, y2), we change all the elements in the rectangle by using "not" operation (if it is a '0' then change it into '1' otherwise change it into '0'). To maintain the information of the matrix, you are asked to write a program to receive and execute two kinds of instructions.  1. C x1 y1 x2 y2 (1 <= x1 <= x2 <= n, 1 <= y1 <= y2 <= n) changes the matrix by using the rectangle whose upper-left corner is (x1, y1) and lower-right corner is (x2, y2).  2. Q x y (1 <= x, y <= n) querys A[x, y].  Input The first line of the input is an integer X (X <= 10) representing the number of test cases. The following X blocks each represents a test case.  The first line of each block contains two numbers N and T (2 <= N <= 1000, 1 <= T <= 50000) representing the size of the matrix and the number of the instructions. The following T lines each represents an instruction having the format "Q x y" or "C x1 y1 x2 y2", which has been described above.  Output For each querying output one line, which has an integer representing A[x, y].  There is a blank line between every two continuous test cases.  Sample Input 1 2 10 C 2 1 2 2 Q 2 2 C 2 1 2 1 Q 1 1 C 1 1 2 1 C 1 2 1 2 C 1 1 2 2 Q 1 1 C 1 1 2 1 Q 2 1 Sample Output 1 0 0 1 Source POJ Monthly,Lou Tiancheng

很经典的二维树状数组题目，网上有很多优秀的文章。查询点转化为查询区间的题目。

一维的类似题目

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <algorithm>
using namespace std;
#define  LL long long
const int N = 1000;
int n,Q;
int sum[N+3][N+3];
int lowbit(int x)
{
    return x&-x;
}
void add(int x,int y)
{
    for(int i=x;i<=n;i+=lowbit(i))
    {
        for(int j=y;j<=n;j+=lowbit(j))
        {
            sum[i][j]++;
        }
    }
}
int query(int x,int y)
{
    int ans=0;
    for(int i=x;i>=1;i-=lowbit(i))
    {
        for(int j=y;j>=1;j-=lowbit(j))
        {
            ans+=sum[i][j];
        }
    }
    return ans;
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d %d",&n,&Q);
        memset(sum,0,sizeof sum);
        while(Q--)
        {
            char c;
            
            cin>>c;
            if(c=='C')
            {
                int x1,x2,y1,y2;
                scanf("%d %d %d %d",&x1,&y1,&x2,&y2);
                add(x1,y1);
                add(x2+1,y2+1);
                add(x2+1,y1);
                add(x1,y2+1);
            }
            else
            {
                int x,y;
                scanf("%d %d",&x,&y);
                int ans=query(x,y);
                printf("%d\n",ans%2);
            }
        }
        if(T) printf("\n");
    }
    
}