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 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

/* Author : yan
 * Question : POJ 2386 Lake Counting
 * Data && Time : Thursday, December 23 2010 10:26 AM
*/
#include
#define bool _Bool
#define true 1
#define false 0
#define MAXN 103
char 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(a1) DFS(a-1,b);
	if(a1 && b1) DFS(a+1,b-1);
	if(b>1) DFS(a,b-1);
	if(b