POJ-2386Lake Counting，搜索题。。

Lake Counting

Time Limit: 1000MS		Memory Limit: 65536K

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.

Source

USACO 2004 November

这应该是搜索里面的简单题了吧，和NYOJ水池数目一模一样，刚学搜索时还不会做，然后看到大牛写的博客上说这是最简单的搜索题了。。。

思路就是用深搜，其实广搜深搜都可以，往八个方向找，只要是W则继续搜索并将该点变为‘.’，做法很多种，看代码；

#include<cstdio>
#include<cmath>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
//const int INF=0x3f3f3f3f;
const int N=150;
char a[N][N];
int n,m;
int dx[8]= {0,0,1,-1,1,1,-1,-1};
int dy[8]= {1,-1,0,0,-1,1,1,-1};
void dfs(int x,int y)
{
    if(x<n&&x>=0&&y>=0&&y<m)
    {
        a[x][y]='.';
        for(int i=0; i<8; i++)
        {
            int xx=x+dx[i];
            int yy=y+dy[i];
            if(a[xx][yy]=='W')
                dfs(xx,yy);
        }
    }
}
int main()
{
    int i,j,sum;
    while(~scanf("%d%d",&n,&m))
    {
        for(i=0; i<n; i++)
            scanf("%s",a[i]);
        sum=0;
        for(i=0; i<n; i++)
            for(j=0; j<m; j++)
                if(a[i][j]=='W')
                {
                    dfs(i,j);
                    sum++;
                }
        printf("%d\n",sum);
    }
    return 0;
}

搜索一定要注意的是边界问题，还有是否会陷入死循环；