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

3

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=105;
int n,m,res;char mp[maxn][maxn];
void dfs(int x,int y){
    mp[x][y]='.';
    for(int dx=-1;dx<=1;++dx){
        for(int dy=-1;dy<=1;++dy){
            int nx=x+dx,ny=y+dy;
            if(0<=nx && nx<n && 0<=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=0;i<n;++i)scanf("%s",mp[i]);
        res=0;
        for(int i=0;i<n;++i)
            for(int j=0;j<m;++j)
                if(mp[i][j]=='W'){dfs(i,j);res++;}
        printf("%d\n",res);
    }
    return 0;
}