poj2386（简单dfs）

就是求图中有多少个水洼。对图进行dfs遍历，并把是水洼的地方全部标记。然后从下一个是水哇的地方再进行dfs。

 1 #include <cstdio>
 2 #include <iostream>
 3 #include <cstring>
 4 using namespace std;
 5 int graph[105][105];
 6 int sum;
 7 bool vis[105][105];
 8 int dx[] = {-1, -1, -1,  0, 0,  1, 1, 1};
 9 int dy[] = {-1,  0,  1, -1, 1, -1, 0 ,1};
10 int row, col;
11
12 bool check(int x, int y)
13 {
14     if( x >= 0 && x < row && y >= 0 && y < col)
15         return true;
16     return false;
17 }
18
19 void dfs(int x, int y)
20 {
21     int ix, iy;
22     for(int i = 0; i < 8; i++)
23     {
24         ix = x + dx[i];
25         iy = y + dy[i];
26         if(!vis[ix][iy] && check(ix, iy) && graph[ix][iy] == 1)
27         {
28             vis[ix][iy] = true;
29             dfs(ix, iy);
30         }
31     }
32 }
33 int main()
34 {
35     //freopen("in.txt", "r", stdin);
36     cin >> row >> col;
37     getchar();
38     memset(vis, false, sizeof(vis));
39
40     char c;
41     for(int i = 0; i < row; i ++)
42     {
43         for(int j = 0; j < col; j++)
44         {
45             c = getchar();
46             graph[i][j] = (c == ‘.‘) ? 0 : 1;
47         }
48         getchar();
49     }
50     sum = 0;
51
52     for(int i  = 0; i < row; i ++)
53     {
54         for(int j = 0; j < col; j++)
55         {
56             if(graph[i][j] == 1 && vis[i][j] == false)
57             {
58                 sum++;
59                 vis[i][j] = true;
60                 dfs(i, j);
61             }
62         }
63     }
64
65     cout << sum << endl;
66     return 0;
67 }