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最新推荐文章于 2022-07-24 19:58:51 发布

Never_Say-Never

最新推荐文章于 2022-07-24 19:58:51 发布

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分类专栏： DFS

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本文介绍了一个算法问题，即计算二维网格中形成的独立水洼数量。通过深度优先搜索(DFS)遍历每个单元格，标记已访问过的水域，最终确定总的水洼数。此问题考察了图遍历和矩阵处理能力。

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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

Hint

OUTPUT DETAILS:

There are three ponds: one in the upper left, one in the lower left,and one along the right side.

题意：有一个大小为N*M的院子，雨后积起了水。八连通的积水被认为连接在一起，（即属于一个水洼）。求院子共有多少水洼.

代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int n,m,ans,nx,ny;
char map[110][110];
int used[110][110];
void dfs(int x,int y)
{
	map[x][y]='.';//这里直接将其现在所在位置替换为'.'。以免下次重复搜索 
	for(int dx=-1;dx<=1;dx++)
	for(int dy=-1;dy<=1;dy++){
		int nx=x+dx,ny=y+dy;
		if(nx>=0&&nx<n&&ny>=0&&ny<m&&map[nx][ny]=='W')
		{

		dfs(nx,ny);	
		}
		
	}
	return ;
}
int main()
{
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		int i,j;
		memset(used,0,sizeof(used));
		ans=0;
		for(i=0;i<n;i++)
		scanf("%s",map[i]);
		for(i=0;i<n;i++)
		{
			for(j=0;j<m;j++)
			{
				if(map[i][j]=='W')
				{
				
					dfs(i,j);
					ans++;
				}
			}
		}
		printf("%d\n",ans);
	}
}