Poj 2386 Lake Counting

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                                                       ***Lake Counting***

Time Limit: 1000MS  Memory Limit: 65536K
Total Submissions: 24201  Accepted: 12216 

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

解题思路：
dfs  ，注意从8个方面进行搜索：
上代码：

#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

int n,m,ans;
char map[110][110];

bool judge(int i,int j)
{
    if(i>=n || i<0)
        return false;
    if(j>=m || j<0)
        return false;
    if(map[i][j] == '.')
        return false;
    return true;
}

void dfs(int i,int j)
{
    map[i][j] = '.';
    if(judge(i+1, j))
        dfs(i+1, j);
    if(judge(i, j+1))
        dfs(i, j+1);
    if(judge(i-1, j))
        dfs(i-1, j);
    if(judge(i, j-1))
        dfs(i, j-1);
    if(judge(i+1, j+1))
        dfs(i+1, j+1);
    if(judge(i-1, j-1))
        dfs(i-1, j-1);
    if(judge(i+1, j-1))
        dfs(i+1, j-1);
    if(judge(i-1, j+1))
        dfs(i-1, j+1);
}

int main()
{
    while(cin>>n>>m)
    {
        getchar();
        ans=0;
        for(int i=0; i<n; i++)
            gets(map[i]);
        for(int i=0; i<n; i++)
        {
            for(int j=0; j<m; j++)
            {
                if(map[i][j] == 'W')
                {
                    ans++;
                    dfs(i,j);
                }
            }
        }
        cout<<ans<<endl;
    }
    return 0;
}