POJ 2155 Matrix

Matrix

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 14330		Accepted: 5411

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
题目大意：输入为C则输入左上角和右下角坐标，然后将这两点中间的矩阵的所有元素取反，输入为Q则输入某一点的坐标，并求出对应坐标点的值

#include 
#include 
#include 
using namespace std;

#define  MAXNUM  1005
int t[MAXNUM][MAXNUM];
int N;

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

void add(int x, int y, int num)
{
    for (int i = x; i > 0; i -= lowbit(i))
    {
        for (int j = y; j > 0; j -= lowbit(j))
        {
            t[i][j] ^= num;
        }
    }
}

int getsum(int x, int y)
{
    int sum = 0;
    for (int i = x; i <= N; i += lowbit(i))
    {
        for (int j = y; j <= N; j += lowbit(j))
        {
            sum ^= t[i][j];
        }
    }
    return sum;
}

int main()
{
    int ncase;
    int M;
    scanf("%d", &ncase);
    while(ncase--)
    {
        memset(t, 0, sizeof(t));
        scanf("%d%d", &N, &M);
        while(M--)
        {
            char c;
            getchar();
            scanf("%c", &c);
            if (c == 'C')
            {
                int x1, y1, x2, y2;
                scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
                add(x2, y2, 1);
                add(x1 - 1, y2, 1);
                add(x2, y1 - 1, 1);
                add(x1 - 1, y1 - 1, 1);
            }
            else
            {
                int x ,y;
                scanf("%d%d", &x, &y);
                printf("%d\n", getsum(x, y));
            }
        }
        printf("\n");
    }
    return 0;
}

 

转载于:https://www.cnblogs.com/lzmfywz/archive/2013/06/02/3113585.html