POJ 2386 Lake Counting 简单的DFS搜索

 

http://poj.org/problem?id=2386

简单DFS搜索

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 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

/* Author : yan * Question : POJ 2386 Lake Counting * Data && Time : Thursday, December 23 2010 10:26 AM*/#include<stdio.h>#define bool _Bool#define true 1#define false 0#define MAXN 103char value[MAXN][MAXN];bool visited[MAXN][MAXN];int n,m;int cnt;void DFS(int a,int b){if(visited[a][b]==true || value[a][b]=='.') return;visited[a][b]=true;if(a>1 && b>1) DFS(a-1,b-1);if(a<n && b<m) DFS(a+1,b+1);if(a>1) DFS(a-1,b);if(a<n) DFS(a+1,b);if(a>1 && b<m) DFS(a-1,b+1);if(a<n && b>1) DFS(a+1,b-1);if(b>1) DFS(a,b-1);if(b<m) DFS(a,b+1);}int main(){//freopen("input","r",stdin);int i,j;int ans=-1;scanf("%d %d",&n,&m);for(i=1;i<=n;i++){for(j=1;j<=m;j++){scanf(" %c",&value[i][j]);}}cnt=0;for(i=1;i<=n;i++){for(j=1;j<=m;j++){if( !visited[i][j] && value[i][j]=='W' ){DFS(i,j);cnt++;}//if(ans<cnt) ans=cnt;//printf("%d/n",cnt);}}printf("%d",cnt);return 0;}