UVa 572 油田（DFS求连通块）

https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=513

终于开始接触图了，恩，开始接触DFS了，这道题就是求连通分量，比较简单。

 1 #include<iostream>
 2 #include<cstring>
 3 using namespace std;
 4
 5 int m, n;                                               //记录连通块的数量
 6 char a[10000][10000];
 7 int  b[10000][10000];
 8
 9 void DFS(int i, int j, int count)
10 {
11     if (i < 0 || i >= m || j < 0 || j >= n) return;     //在递归时出界的情况
12         if(b[i][j]>0 || a[i][j]!=‘@‘)  return;
13     b[i][j] = count;
14     for (int p = -1; p <= 1; p++)
15     {
16         for (int q = -1; q <= 1; q++)
17         {
18             if (p != 0||q != 0)  DFS(i + p, j + q, count);
19         }
20     }
21 }
22
23 int main()
24 {
25     while (cin >> m >> n && m && n)
26     {
27         for (int i = 0; i < m; i++)
28             scanf("%s", &a[i]);
29         int count = 0;
30         memset(b, 0, sizeof(b));
31         for (int i = 0; i < m; i++)
32         {
33             for (int j = 0; j < n; j++)
34             {
35                 if (b[i][j] == 0 && a[i][j]==‘@‘) DFS(i, j, ++count);
36             }
37         }
38         cout << count << endl;
39     }
40     return 0;
41 }

2016-12-03  07:49:56