pku 3363(内部测试赛)

Annoying painting tool

Time Limit: 1000MS Memory limit: 65536K

题目描述

Maybe you wonder what an annoying painting tool is? First of all, the painting tool we speak of supports only black and white. Therefore, a picture consists of a rectangular area of pixels, which are either black or white. Second, there is only one operation how to change the colour of pixels:

Select a rectangular area of r rows and c columns of pixels, which is completely inside the picture. As a result of the operation, each pixel inside the selected rectangle changes its colour (from black to white, or from white to black).

Initially, all pixels are white. To create a picture, the operation described above can be applied several times. Can you paint a certain picture which you have in mind?

输入

The input contains several test cases. Each test case starts with one line containing four integers n, m, r and c. (1 ≤ r ≤ n ≤ 100, 1 ≤ c ≤ m ≤ 100), The following nlines each describe one row of pixels of the painting you want to create. The i^th line consists of m characters describing the desired pixel values of the i^th row in the finished painting (\'0\' indicates white, \'1\' indicates black).

The last test case is followed by a line containing four zeros.

输出

For each test case, print the minimum number of operations needed to create the painting, or -1 if it is impossible.

示例输入

3 3 1 1
010
101
010
4 3 2 1
011
110
011
110
3 4 2 2
0110
0111
0000
0 0 0 0

示例输出

4
6
-1

将每一个1看成小矩形的左上角  ，每一个点（左上角）只能被反转一次

#include<stdio.h>
#define N 150
char map[N][N];
int main()
{
    int n,m,r,c,i,j,k,l;
    while(scanf("%d%d%d%d",&n,&m,&r,&c),n+m+r+c)
    {
        int num=0;
        getchar();
        int f=0;
        for(i=0;i<n;i++)scanf("%s",map[i]);
        for(i=0;i<n;i++)
        {
            for(j=0;j<m;j++)
            {
                if(map[i][j]=='1')
                {
                    if(i+r-1>=n||j+c-1>=m)break;
                    num++;
                    for(k=i;k<i+r;k++)
                    {
                        for(l=j;l<j+c;l++)
                        {
                            if(map[k][l]=='1')map[k][l]='0';
                            else map[k][l]='1';
                        }
                    }
                }
            }
            if(j<m)f=1;
        }
        if(f)printf("-1\n");
        else
        printf("%d\n",num);
    }
}