POJ 2386 LakeCounting

Lake Counting

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 19975		Accepted: 10048

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.

算法分析：DFS深度搜索问题，树形搜索。代码如下：

#include<stdio.h>
#define MAXN 101
int N,M;
char a[MAXN][MAXN];
void dfs(int x,int y);
int main()
{
	int res=0,i,j;
	scanf("%d%d",&N,&M);
	getchar();
	for(i=0;i<N;i++)
	{
		for(j=0;j<M;j++)
			scanf("%c",&a[i][j]);
		getchar();
	}
	for(i=0;i<N;i++)
	{
		for(j=0;j<M;j++)
		{
			if(a[i][j]=='W')
			{
				dfs(i,j);res++;
			}
		}
	}
	printf("%d\n",res);
	return 0;
}
void dfs(int x,int y)
{
	int dx,dy,nx,ny;
	a[x][y]='.';
	for(dx=-1;dx<=1;dx++)
	{
		for(dy=-1;dy<=1;dy++)
		{
			nx=dx+x;ny=dy+y;
			if(0<=nx&&nx<N&&0<=ny&&ny<M&&a[nx][ny]=='W')
			{
				dfs(nx,ny);
			}
		}
	}
	return;	
}