Lake Counting(poj 2386)

题目描述：

Description

Due to recent rains, water has pooled in various places in Farmer John's field, which is represented by a rectangle of N x M (1 <= N <= 100; 1 <= M <= 100) squares. Each square contains either water ('W') or dry land ('.'). Farmer John would like to figure out how many ponds have formed in his field. A pond is a connected set of squares with water in them, where a square is considered adjacent to all eight of its neighbors.

Given a diagram of Farmer John's field, determine how many ponds he has.

Input

* Line 1: Two space-separated integers: N and M

* Lines 2..N+1: M characters per line representing one row of Farmer John's field. Each character is either 'W' or '.'. The characters do not have spaces between them.

Output

* Line 1: The number of ponds in Farmer John's field.

Sample Input

10 12
W........WW.
.WWW.....WWW
....WW...WW.
.........WW.
.........W..
..W......W..
.W.W.....WW.
W.W.W.....W.
.W.W......W.
..W.......W.

Sample Output

3
代码如下：

 #include<iostream>
 char map[][];
 int n,m;
 void dfs(int i,int j);
 int main()
 {
     using namespace std;
     //int n,m;
     int sum = ;
     cin >> n >> m;
     //char map[n][m];
     for(int i = ;i < n;i++)
         for(int j = ;j < m;j++)
             cin >> map[i][j];
     for(int i = ;i < n;i++)
         for(int j = ;j < m;j++)
         {
             if(map[i][j] == 'W')
             {
                 dfs(i,j);
                 sum++;
             }
         }
     cout << sum << endl;
     return ;
 }

 void dfs(int i,int j)
 {
     int x,y;
     map[i][j] = '.';
     for(int nx = -;nx <= ;nx++)
         for(int ny = -;ny <= ;ny++)
         {
             x = i + nx;
             y = j + ny;
             if(x <= n && y <= m && x >=  && y >=  && map[x][y] == 'W')
                 dfs(x,y);
         }
 }

代码分析：

这道题用到了深度优先搜索法，对这个算法也是最近刚接触，所以可能说的不太好，所以恳请读者指正。

深度优先搜索法，通俗地讲，就是从初始状态，在一个方向上，一直访问到最后一个状态，然后再返回到前一个个状态，换个一个方向，继续访问到最后一个状态。

在这道题目上，我们从任意一个的'W'可以访问，看它的周围的八个状态，哪个状态是'W'，就从这个状态再继续深入，直到某个状态周围的八个状态都不是'W'，则返回前一个状态，直到这个方向上都不是'W'，则这个方向的状态都访问完。在这里我们将'W'改为'.'，表示访问过。

1次dfs后与初始的这个W连接的所以W就都被替换为'.'，直到图中不再存在W为止，总共进行dfs的次数就是答案。

参考书籍：［挑战程序设计竞赛］