Lake Counting(POJ 2386)

Lake Counting

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 44751		Accepted: 22120

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

Hint

OUTPUT DETAILS: 

There are three ponds: one in the upper left, one in the lower left,and one along the right side.

 

• 题解：从任意的'W'开始，不停地把邻接的部分用'.'代替，1次dfs后与初始的这个W连接的所有'W'就都被替换成了'.'，因此直到图中不再存在w为止，总共进行dfs的次数就是答案了，复杂度是O(n*m)
• 代码：

  #include <iostream>
  
  using namespace std;
  
  int n,m;
  char ** a;
  
  void dfs(int, int);
  
  int main()
  {
      cin >> n >> m;
      a = new char*[n];
      for (int i=0; i<n; i++) a[i]=new char[m];
      for (int i=0; i<n; i++)
      {
          for (int j=0; j<m; j++)
          {
              cin >> j[i[a]];
          }
      }
      int s=0;
      for (int i=0; i<n; i++)
      {
          for (int j=0; j<m; j++)
          {
              if (j[i[a]]=='W') 
              {
                  dfs(i,j);
                  ++s;
              }
          }
      }
      cout << s << endl;
  }
  
  void dfs(int x, int y)
  {
      y[x[a]] ='.';
      for (int dx=-1; dx<=1; dx++)
      {
          for (int dy=-1; dy<=1; dy++)
          {
              int xx=x+dx, yy=y+dy;
              if (xx>=0 && xx<n && yy>=0 && yy<m && yy[xx[a]]=='W') dfs(xx,yy);
          }
      }
  }