poj 2386 Lake Counting

最新推荐文章于 2022-07-24 19:58:51 发布

蕾姆的小迷弟

最新推荐文章于 2022-07-24 19:58:51 发布

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分类专栏： dfs poj 文章标签： dfs

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本文介绍了一种使用深度优先搜索（DFS）算法解决湖泊计数问题的方法。问题设定在一个由N*M网格组成的矩形区域中，每个格子代表水或干燥土地。目标是通过DFS遍历来确定该区域中形成的独立湖泊数量。

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Lake Counting

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 41379		Accepted: 20528

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

统计大水坑个数，W为小水坑，用dfs来做.

#include <iostream>
using namespace std;
char a[105][105];
void dfs(int x,int y,int n,int m)
{
	a[x][y]='.';                   //先把此点转化掉，然后开始寻找符合条件范围的点，找到后开始递归搜索 
	int dx,dy;
	for(dx=-1; dx<=1; dx++)
	{
		for(dy=-1; dy<=1; dy++)
		{
			int nx=x+dx,ny=y+dy;
			if(a[nx][ny]=='W' && nx<n && ny<m && nx>=0 && ny>=0)
			{
				dfs(nx,ny,n,m);
			}
		}
	}
	return ;
}
int main()
{
	int n,i,j,m;
	cin>>n>>m;
	for(i=0; i<n; ++i)
	{
		for(j=0; j<m; ++j)
		{
			cin>>a[i][j];
		}
	}
	int ans=0;
	for(i=0; i<n; ++i)
	{
		for(j=0; j<m; ++j)
		{
			if(a[i][j]=='W')  //从头开始找到W则开始深搜 
			{
				dfs(i,j,n,m);
				ans++;
			}
		}
	}
	cout<<ans<<endl;
	return 0;   
}