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poj2386 Lake Counting（简单DFS）

2024-07-12 05:33:58   219人阅读

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题目链接：http://poj.org/problem?id=1562


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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
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;
}
































        

            
                
                  
                
                

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