POJ-2386

Lake Counting

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 36208		Accepted: 17982

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.

题意：

八连通的'W'为一片水洼，问总共有多少片水洼。

对于每个'W'，搜索其八连通区域，将所有找到的'W'都变成'.'，直至所有点都变成'.'。

期间所进行dfs的次数就是水洼的数目。

AC代码：

#include<bits/stdc++.h>//POJ提交注意更换头文件
using namespace std;

char mp[110][110];

int n,m;

void dfs(int i,int j){
    mp[i][j]='.';
    for(int dx=-1;dx<=1;dx++){
        for(int dy=-1;dy<=1;dy++){
            int nx=i+dx,ny=j+dy;
            if(0<=nx&&nx<n&&0<=ny&&ny<m&&mp[nx][ny]=='W')
            dfs(nx,ny);
        }
    }
    return ;
}

int main(){
    ios::sync_with_stdio(false);
    while(cin>>n>>m){
        int ans=0;
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                cin>>mp[i][j];
            }
        }
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                if(mp[i][j]=='W'){
                    dfs(i,j);
                    ans++;
                }
            }
        }
        cout<<ans<<endl;
    }
    return 0;
}