POJ1681-Painter's Problem[十字枚举]

Painter's Problem

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 6306		Accepted: 3006

Description

There is a square wall which is made of n*n small square bricks. Some bricks are white while some bricks are yellow. Bob is a painter and he wants to paint all the bricks yellow. But there is something wrong with Bob's brush. Once he uses this brush to paint brick (i, j), the bricks at (i, j), (i-1, j), (i+1, j), (i, j-1) and (i, j+1) all change their color. Your task is to find the minimum number of bricks Bob should paint in order to make all the bricks yellow. 

Input

The first line contains a single integer t (1 <= t <= 20) that indicates the number of test cases. Then follow the t cases. Each test case begins with a line contains an integer n (1 <= n <= 15), representing the size of wall. The next n lines represent the original wall. Each line contains n characters. The j-th character of the i-th line figures out the color of brick at position (i, j). We use a 'w' to express a white brick while a 'y' to express a yellow brick.

Output

For each case, output a line contains the minimum number of bricks Bob should paint. If Bob can't paint all the bricks yellow, print 'inf'.

Sample Input

2
3
yyy
yyy
yyy
5
wwwww
wwwww
wwwww
wwwww
wwwww

Sample Output

0
15

思路：定义矩阵color[][]表示正方形的颜色，0表示黄色，1表示白色。定义矩阵paint[][]，paint[i][j]为1表示格子(i,j)涂色，否则为0表示未涂色。然后枚举paint[1]整行，接下来根据(i-1,j)格子的颜色决定(i,j)格子是否涂色。最终，

判断最后一行涂色是否满足要求正方形的最后一行。

AC代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int MAXN = 20;
const int INF = 0x3f3f3f3f;
int color[MAXN][MAXN];
int paint[MAXN][MAXN];
int n;
int judge() {
    for (int i = 1; i < n; i++) {
        for (int j = 1; j <= n; j++) {
            paint[i + 1][j] = (color[i][j] + paint[i][j] + paint[i - 1][j] + paint[i][j - 1] + paint[i][j + 1]) % 2;    
            //根据上一格的颜色判断本格是否涂色
        }
    }
    for (int i = 1; i <= n; i++) {
        if ((paint[n][i] + paint[n][i - 1] + paint[n][i + 1] + paint[n - 1][i]) % 2 != color[n][i]) {   //判断最后一行是否满足
            return INF;
        }
    }
    int ret = 0;
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= n; j++) {
            if (paint[i][j] == 1) {
                ret++;
            }
        }
    }
    return ret;
}
int main() {
    int T;
    scanf("%d", &T);
    while (T--) {
        scanf("%d", &n);
        memset(paint, 0, sizeof(paint));
        for (int i = 1; i <= n; i++) {
            char s[MAXN];
            scanf("%s", s);
            for (int j = 0; s[j]; j++) {
                if (s[j] == 'w') {
                    color[i][j + 1] = 1;
                } else {
                    color[i][j + 1] = 0;

                }
            }
        }

        int res = INF;
        for (int i = 0; i < (1 << n); i++) {
            for (int k = 0; k < n; k++) {
                if (i & (1 << k)) {
                    paint[1][k + 1] = 1;    //枚举第一行
                } else {
                    paint[1][k + 1] = 0;
                }
            }
            res = min(res, judge());
        }
        if (res == INF) {
            printf("inf\n");
        } else {
            printf("%d\n", res);
        }
    }
    return 0;
}