推荐：POJ2386 Lake Counting 【DFS】

Lake Counting Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 20782   Accepted: 10473 Description Due to recent rains, water has pooled i

2018阿里云全部产品优惠券(新购或升级都可以使用，强烈推荐)
领取地址：https://promotion.aliyun.com/ntms/yunparter/invite.html?userCode=zo93kaue

转载请注明出处：http://blog.csdn.net/u012860063?viewmode=contents

题目链接：http://poj.org/problem?id=1562

Description

推荐：PPOJ 2386 Lake Counting（DFS）

题意：有一块N*M的土地，有些地方有水，有些地方很干燥，有水标记为'W'，干燥为'.'。八连通的积水被认为是连接在一起的(即两个都在同一个九宫格内)，问一共有多

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

代码如下：

#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;
}

推荐：POJ 2386 Lake Counting(dfs or bfs)

Lake Counting Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 24999 Accepted: 12619 Description Due to recent rains, water has pooled in va