sdut oj 2372 Annoying painting tool (【暴力枚举测试】1Y )

Annoying painting tool

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

代码：

#include <iostream>
#include <string>
#include <algorithm>
#include <stdio.h>
#include <string.h>

using namespace std;

int map[110][110];
int n, m, r, c;
bool judge(int dd, int ff)
{
    if((dd+r-1)<=n && (ff+c-1)<=m )
        return true;
    else
        return false;
}

void OP(int dd, int ff)
{
    int i, j;
    for(i=dd; i<=(dd+r-1); i++)
    {
        for(j=ff; j<=(ff+c-1); j++)
        {
            if(map[i][j]==1)
                map[i][j]=0;
            else
                map[i][j]=1;
        }
    }
}

int main()
{
    int flag;
    int i, j, k, e;
    char s[110];
    int cnt;
    while(scanf("%d %d %d %d", &n, &m, &r, &c)!=EOF)
    {
        if(n==0&&m==0&&r==0&&c==0 )
            break;
        flag=1;
        cnt=0;
        for(i=1; i<=n; i++)
        {
            scanf("%s", s);
            int len=strlen(s);
            for(j=0; j<len; j++)
                map[i][j+1]=s[j]-48;
        }
        for(i=1; i<=n; i++)
        {
            for(j=1; j<=m; j++)
            {
                if(map[i][j]==1)
                {
                    if(judge(i, j)==true)
                    {
                        OP(i, j); cnt++;
                    }
                    else
                    {
                        flag=0; break;
                    }
                }
            }
            if(flag==0) break;
        }
        if(flag==0)
            printf("-1\n");
        else
            printf("%d\n", cnt );
    }
    return 0;
}