POJ-3363 Annoying painting tool

Annoying painting tool

Time Limit: 1000MS Memory Limit: 65536K

Description

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?

Input

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

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

Output

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

Sample Input

3 3 1 10101010104 3 2 10111100111103 4 2 20110011100000 0 0 0

Sample Output

46-1

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前言：许久之前的代码了，但是不知道为什么没有发题解。。。补上

思路：贪心。给定的刷子会把某区域内的颜色全部反转，因为刷两下和不刷是一模一样的，所以从左上角开始，一旦遇到了1，就刷一下。一个一个枚举，到最后如果图上还存在1，那就是不可能消除的。

代码如下：

/****************************************/ #include <cstdio> #include <cstdlib> #include <cstring> #include <algorithm> #include <cmath> #include <stack> #include <queue> #include <vector> #include <map> #include <string> #include <iostream> using namespace std;/****************************************/bool mat[105][105];//bool mat[105][105];int main(){int n, m, r, c;while(~scanf("%d%d%d%d", &n, &m, &r, &c), n||m||r||c) {char t;for(int i = 0; i < n; i++) {for(int j = 0; j < m; j++) {scanf(" %c", &t);mat[i][j] = t - '0';}}int op = 0;bool ok = true;for(int i = 0; i+r <= n; i++) {for(int j = 0; j+c <= m; j++) {if(mat[i][j] == 1) {op++;for(int k = i; k < i+r; k++) {for(int kk = j; kk < j+c; kk++) {mat[k][kk] = !mat[k][kk];}}}}}for(int i = 0; i < n; i++) {for(int j = 0; j < m; j++) {if(mat[i][j] == 1) {ok = false;break;}}}printf("%d\n", ok ? op : -1);}return 0;}