POJ 2155 Matrix【二维树状数组】POJ 2155【

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


思路:二维树状数组快速求数字子矩阵的和.
代码如下：

#include<stdio.h>
#include<string.h>
int c[1005][1005], n; 
int Lowbit(int t)
{
    return t&(-t);
}
void add(int x, int y, int val)
{
    int i=y;
    while(x<=n)
    {
        y=i;
        while(y<=n)
        {
            c[x][y]+=val;
            y+=Lowbit(y);
        }
        x+=Lowbit(x);
    }
}
int Sum(int x, int y)
{
    int i=y, sum=0;
    while(x>0)
    {
        y=i;
        while(y>0)
        {
            sum+=c[x][y];
            y-=Lowbit(y);
        }
        x-=Lowbit(x);
    }
    return sum;
}
int main()
{
    int T, m, x1, y1, x2, y2, i;
    char ch; 
    scanf("%d", &T);
    while(T--)
    {
        scanf("%d%d", &n, &m);
        memset(c, 0, sizeof(c)); 
        getchar();
        for(i=0; i<m; i++)
        {
            scanf("%c", &ch); 
            if(ch=='C')
            {
                scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
                x2++, y2++; 
                add(x1, y1, 1);
                add(x2, y2, 1);
                add(x1, y2, -1);
                add(x2, y1, -1);
            }
            else
            {
                scanf("%d%d", &x1, &y1);
                printf("%d\n", Sum(x1, y1)%2);
            }
            getchar(); 
        }
        printf("\n"); 
    }
}