poj 2386 Lake Counting

                                                                                                                                  Lake Counting

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 14703		Accepted: 7452

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

题意： 判断w相连的区域共有多少个……

题解：很早之前做的水题，今天又打了一遍……

代码：

#include<iostream>
using namespace std;
char a[1000][1000];
bool vis[1000][1000];
int dfs(int i,int j)
{
    if (a[i][j]=='.')
        return 0;
    if (a[i][j]=='W'&&!vis[i][j])
    {

        vis[i][j]=1;
        dfs(i+1,j);
        dfs(i-1,j);
        dfs(i,j+1);
        dfs(i,j-1);
        dfs(i+1,j+1);
        dfs(i-1,j-1);
        dfs(i-1,j+1);
        dfs(i+1,j-1);
        return 1;
    }
    return 0;


}
int main()
{
    int ans=0;
    int n,m;
    cin>>n>>m;
    for (int i=0;i<n;i++)
    {
        for (int j=0;j<m;j++)
        {
            cin>>a[i][j];
            vis[i][j]=0;
        }
    }
    for (int i=0;i<n;i++)
    {
        for (int j=0;j<m;j++)
        {

            if (dfs(i,j)) ans++;
        }
    }
    cout<<ans<<endl;
    return 0;
}