2013
12-23

# Annoying painting tool

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?

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 n lines each describe one row of pixels of the painting you want to create. The ith line consists of m characters describing the desired pixel values of the ith row in the finished painting (’0′ indicates white, ’1′ indicates black).

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

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

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

4
6
-1

#include<iostream>
#include<cstring>
using namespace std;
const int maxn=110;
int map[maxn][maxn],color[maxn][maxn];
char s[maxn][maxn];
int n,m,r,c,ans;
void change()
{
int i,j,k,p;
for(i=0;i<n;i++)
{
for(j=0;j<m;j++)
{
if(i+r<=n&&j+c<=m&&map[i][j]!=color[i][j])//限制寻找条件，在边界范围之内，并且与所给的颜色不同
{
for(k=i;k<i+r;k++)
{
for(p=j;p<j+c;p++)
if(color[k][p])
color[k][p]=0;
else
color[k][p]=1;
}
ans++;
}

}
}
}
int judge()//判断最后是否与所给的颜色完全相同
{
int i,j,k;
for(i=0;i<n;i++)
{
for(j=0;j<m;j++)
{
if(map[i][j]!=color[i][j])
return 0;
}
}
return 1;
}

int main(void)
{
int i,j;
while(scanf("%d%d%d%d",&n,&m,&r,&c)!=EOF)
{
if(n==0&&m==0&&r==0&&c==0)
break;
memset(map,0,sizeof(map));//对其进行初始化
memset(color,0,sizeof(color));
memset(s,0,sizeof(s));
ans=0;
for(i=0;i<n;i++)
{
scanf("%s",s[i]);
for(j=0;j<m;j++)
{
map[i][j]=s[i][j]-'0';
}
}
change();
if(judge())
printf("%d\n",ans);
else
printf("-1\n");
}
return 0;
}

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