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：

View Code

#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"); 

    }

}