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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 nlines 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 <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;
}
sdut oj 2372 Annoying painting tool (【暴力枚举测试】1Y )
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原文地址：http://www.cnblogs.com/yspworld/p/4266730.html
