Enum:Fliptile(POJ 3279)

 1 #include <iostream>
 2 #include <functional>
 3 #include <algorithm>
 4 
 5 using namespace std;
 6 
 7 static int map[18][18], flip[18][18], ans_map[18][18], M, N;
 8 static int dirx[4] = { -1, 0, 0, 1 };
 9 static int diry[4] = { 0, -1, 0, 0 };
10 
11 int solve(void);
12 int get_color(const int, const int);
13 
14 int main(void)//开关问题，枚举第一行即可得到下面几行的关系
15 {
16     int ans, tmp;
17     while (~scanf("%d%d", &N, &M))//读图
18     {
19         for (int i = 0; i < N; i++)
20             for (int j = 0; j < M; j++)
21                 scanf("%d", &map[i][j]);
22         ans = INT_MAX;
23         for (int i = 0; i < 1 << M; i++)//枚举第一行
24         {
25             memset(flip, 0, sizeof(flip));
26             for (int j = 0; j < M; j++)//枚举结果填入第一行
27                 if ((1 << j & i) != 0)
28                     flip[0][M - j - 1] = 1;
29             tmp = solve();
30             if (tmp >= 0 && tmp < ans)
31             {
32                 ans = tmp;
33                 memcpy(ans_map, flip, sizeof(flip));
34             }
35         }
36 
37         if (ans == INT_MAX)
38             printf("IMPOSSIBLE\n");
39         else
40         {
41             for (int i = 0; i < N; i++)
42             {
43                 for (int j = 0; j < M; j++)
44                     printf("%d ",ans_map[i][j]);
45                 printf("\n");
46             }
47         }
48     }
49     return 0;
50 }
51 
52 int solve(void)
53 {
54     int ans = 0;
55     for (int i = 1; i < N; i++)//从第二行开始一行一行模拟
56         for (int j = 0; j < M; j++)
57             if (get_color(j, i - 1) == 1)//记住是上一行翻的牌子
58                 flip[i][j] = 1;//说明需要翻
59 
60     for (int j = 0; j < M; j++)//得到最后一行的所有格子的颜色，如果还有黑色，则说明这一步无法把格子全部翻转成白色
61         if (get_color(j, N - 1) == 1)
62             return -1;
63 
64     for (int i = 0; i < N; i++)//统计步数
65         for (int j = 0; j < M; j++)
66             if (flip[i][j])
67                 ans++;
68     return ans;
69 }
70 
71 int get_color(const int x, const int y)
72 {
73     //得到(x,y)位置的颜色
74     int ny, nx, sum = map[y][x];
75 
76     for (int i = 0; i < 4; i++)
77     {
78         nx = x + dirx[i]; ny = y + diry[i];
79         if ((0 <= nx&&nx < M) && (0 <= ny&&ny < N))
80             sum += flip[ny][nx];
81     }
82     return sum % 2;
83 }