poj2386 Lake Counting（简单DFS）

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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

代码例如以下：

#include <iostream>

#include <algorithm>

using namespace std;

#include <cstring>

#define TM 100+17

int N, M;

char map[TM][TM];

bool vis[TM][TM];

int xx[8]={0,1,1,1,0,-1,-1,-1};

int yy[8]={1,1,0,-1,-1,-1,0,1};

void DFS(int x, int y)

{

	vis[x][y] = true;

	for(int i = 0; i < 8; i++)

	{

		int dx = x+xx[i];

		int dy = y+yy[i];

		if(dx>=0&&dx<N&&dy>=0&&dy<M&&!vis[dx][dy]&&map[dx][dy] == 'W')

		{

			vis[dx][dy] = true;

			DFS(dx,dy);

		}

	}

}

int main()

{

	int i, j;

	while(cin>>N>>M)

	{

		int count = 0;

		memset(vis,false,sizeof(vis));

		for(i = 0; i< N; i++)

		{

			cin>>map[i];

		}

		for(i = 0; i < N; i++)

		{

			for(j = 0; j < M; j++)

			{

				if(map[i][j] == 'W' && !vis[i][j])

				{

					count++;

					DFS(i,j);

				}

			}

		}

		cout<<count<<endl;

	}

	return 0;

}

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poj2386 Lake Counting（简单DFS）

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