POJ-2386-Lake Counting

最新推荐文章于 2019-09-06 19:25:48 发布

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本文介绍了一个经典的图论问题——湖泊计数。问题要求在给定的N*M网格中计算水池的数量，每个水池是由相邻的水域组成。通过深度优先搜索（DFS）算法解决该问题，并提供了完整的AC代码。

Lake Counting

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
题意：给定一个N*M的图，求联通块。

题解：联通块模板题，dfs搜索即可。

AC代码：

/*
* @Author: 王文宇
* @Date:   2017-11-27 15:20:31
* @Last Modified by:   王文宇
* @Last Modified time: 2017-11-27 15:35:20
*/
#include<iostream>
using namespace std;
const int maxn = 105;
char a[maxn][maxn];
int n,m,vis[maxn][maxn],ans;
int ax[4]={-1,1,0,0};
int ay[4]={0,0,-1,1};
void dfs(int x,int y,int z)
{
	vis[x][y]=1;
	for(int i=0;i<4;i++)
	{
		for(int j=0;j<4;j++)
		{
			int x1 = x+ax[i];
			int y1 = y+ay[j];
			if(x1<1||x1>n||y1<1||y1>m)continue;
			if(vis[x1][y1]==0&&a[x1][y1]=='W')dfs(x1,y1,z);
		}
	}
	return ;
}
int main(int argc, char const *argv[])
{
	cin>>n>>m;
	for(int i=1;i<=n;i++)
		for(int j=1;j<=m;j++)
		{
			vis[maxn][maxn]=0;
			cin>>a[i][j];
		}
	ans = 0;
	for(int i=1;i<=n;i++)
		for(int j=1;j<=m;j++)
		{
			if(vis[i][j]==0&&a[i][j]=='W')dfs(i,j,ans++);
		}
	cout<<ans<<endl;
	return 0;
}