题解报告：poj 2386 Lake Counting（dfs求最大连通块的个数）

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

Hint

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

解题思路：从任意的'W'开始，不停地把邻接的部分用'.'代替。一次DFS后与初始的这个'W'连接的所有'W'就被替换成了'.'，因此直到图中不再存在'W'为止，总共进行的DFS的次数就是最终答案。

AC代码：

 #include<iostream>

 #include<cstdio>

 using namespace std;

 const int maxn=;

 int n,m,res;char mp[maxn][maxn];

 void dfs(int x,int y){

     mp[x][y]='.';

     for(int dx=-;dx<=;++dx){

         for(int dy=-;dy<=;++dy){

             int nx=x+dx,ny=y+dy;

             if(<=nx && nx<n && <=ny && ny<m && mp[nx][ny]=='W')dfs(nx,ny);//往8个方向寻找'W'的点

         }

     }

     return;

 }

 int main(){

     while(~scanf("%d%d",&n,&m)){

         for(int i=;i<n;++i)scanf("%s",mp[i]);

         res=;

         for(int i=;i<n;++i)

             for(int j=;j<m;++j)

                 if(mp[i][j]=='W'){dfs(i,j);res++;}

         printf("%d\n",res);

     }

     return ;

 }

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