poj2386Lake Counting

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 12W........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：有一个大小为N*M的院子，雨后积起了水。八连通的积水被认为连接在一起，（即属于一个水洼）。求院子共有多少水洼.

代码：

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;int n,m,ans,nx,ny;char map[110][110];int used[110][110];void dfs(int x,int y){map[x][y]='.';//这里直接将其现在所在位置替换为'.'。以免下次重复搜索 for(int dx=-1;dx<=1;dx++)for(int dy=-1;dy<=1;dy++){int nx=x+dx,ny=y+dy;if(nx>=0&&nx<n&&ny>=0&&ny<m&&map[nx][ny]=='W'){dfs(nx,ny);}}return ;}int main(){while(scanf("%d%d",&n,&m)!=EOF){int i,j;memset(used,0,sizeof(used));ans=0;for(i=0;i<n;i++)scanf("%s",map[i]);for(i=0;i<n;i++){for(j=0;j<m;j++){if(map[i][j]=='W'){dfs(i,j);ans++;}}}printf("%d\n",ans);}}