hihocoder offer收割编程练习赛12 C 矩形分割

思路：

模拟，深搜。

实现：

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <string>
 4 using namespace std;
 5
 6 const int dx[4] = { 0, 1, 0, -1 };
 7 const int dy[4] = { -1, 0, 1, 0 };
 8
 9 int n, m, cnt = 0;
10 int a[305][305];
11 bool vis[305][305];
12
13 void dfs(int x, int y)
14 {
15     vis[x][y] = true;
16     for (int i = 0; i < 4; i++)
17     {
18         int nx = x + dx[i];
19         int ny = y + dy[i];
20         if (nx >= 0 && nx < 3 * n && ny >= 0 && ny < 3 * m && !a[nx][ny] && !vis[nx][ny])
21         {
22             dfs(nx, ny);
23         }
24     }
25 }
26
27 int main()
28 {
29     scanf("%d %d", &n, &m);
30     getchar();
31     for (int i = 0; i < n; i++)
32     {
33         for (int j = 0; j < m; j++)
34         {
35             char c = getchar();
36             if (c == ‘/‘)
37             {
38                 a[3 * i][3 * j] = 0;
39                 a[3 * i][3 * j + 1] = 0;
40                 a[3 * i][3 * j + 2] = 1;
41                 a[3 * i + 1][3 * j] = 0;
42                 a[3 * i + 1][3 * j + 1] = 1;
43                 a[3 * i + 1][3 * j + 2] = 0;
44                 a[3 * i + 2][3 * j] = 1;
45                 a[3 * i + 2][3 * j + 1] = 0;
46                 a[3 * i + 2][3 * j + 2] = 0;
47             }
48             else if (c == ‘\\‘)
49             {
50                 a[3 * i][3 * j] = 1;
51                 a[3 * i][3 * j + 1] = 0;
52                 a[3 * i][3 * j + 2] = 0;
53                 a[3 * i + 1][3 * j] = 0;
54                 a[3 * i + 1][3 * j + 1] = 1;
55                 a[3 * i + 1][3 * j + 2] = 0;
56                 a[3 * i + 2][3 * j] = 0;
57                 a[3 * i + 2][3 * j + 1] = 0;
58                 a[3 * i + 2][3 * j + 2] = 1;
59             }
60         }
61         getchar();
62     }
63
64     for (int i = 0; i < 3 * n; i++)
65     {
66         for (int j = 0; j < 3 * m; j++)
67         {
68             if (!a[i][j] && !vis[i][j])
69             {
70                 dfs(i, j);
71                 cnt++;
72             }
73         }
74     }
75     cout << cnt << endl;
76     return 0;
77 }