Satellite Photographs（dfs）

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原创最新推荐文章于 2026-08-17 11:16:49 发布 764 阅读 AI 写同款· GEO 优化 ›

POJ 4103 踩方格（简单的DFS问题） POJ 4103 踩方格思路1：DFS 用二维数组存储走路信息（初试为0）用变量step记录走了几个格子（起点算1，与0区分开） DFS函数分4部分： 1.返回 2.修改 3.递归 4.复原由于本题需要复原，需要注意两点： 1.如果递归访问下一个格子，已经塌了，则不需要复原。 2.注意step–的操作，step++的操作以及step==？的返回操作的相对位置。 DFS代码 void dfs... 阅读详情

Farmer John purchased satellite photos of W x H pixels of his farm (1 <= W <= 80, 1 <= H <= 1000) and wishes to determine the largest 'contiguous' (connected) pasture. Pastures are contiguous when any pair of pixels in a pasture can be connected by traversing adjacent vertical or horizontal pixels that are part of the pasture. (It is easy to create pastures with very strange shapes, even circles that surround other circles.)

Each photo has been digitally enhanced to show pasture area as an asterisk ('*') and non-pasture area as a period ('.'). Here is a 10 x 5 sample satellite photo:

..*.....** .**..***** .*...*.... ..****.*** ..****.***

This photo shows three contiguous pastures of 4, 16, and 6 pixels. Help FJ find the largest contiguous pasture in each of his satellite photos.

Input

* Line 1: Two space-separated integers: W and H

* Lines 2..H+1: Each line contains W "*" or "." characters representing one raster line of a satellite photograph.

Output

* Line 1: The size of the largest contiguous field in the satellite photo.

Sample Input

10 5
..*.....**
.**..*****
.*...*....
..****.***
..****.***

Sample Output

给你一个图，有* 和 .，*是被.隔离成若干块，求这若干块中*数量最多是多少？

#if 0
#include<iostream>
#include<cstring>
using namespace std;
int n,m,maxx,cnt;
char a[1010][1010];
bool vis[1010][1010];
int f[4][2]={{1,0},{-1,0},{0,1},{0,-1}}; 

bool ok(int i,int j)
{
	if(i>=1&&i<=n&&j>=1&&j<=m) 
	{
		
			return 1; 
	}
	
	return 0; 
}


void dfs(int x,int y) 
{
	
		
		if(ok(x,y)==0) 
		{
			return;
		}
		
		if(a[x][y]=='*'&&vis[x][y]==0)
		{
			vis[x][y]=1;
			cnt++;
			
			for(int i=0; i<4; i++)
			{
				int xi=x+f[i][0];
				int yi=y+f[i][1];
				dfs(xi,yi);				
			}
		}
	
	
	
}
	
int main()
{
	while(cin>>m>>n)
	{
		memset(a,0,sizeof(a));
		memset(vis,0,sizeof(vis));
		cnt=0;
		
		for(int i=1; i<=n; i++)
		{
			for(int j=1; j<=m; j++)
			{
				cin>>a[i][j];	
			}
		}
		
		for(int i=1; i<=n; i++)
		{
			for(int j=1; j<=n; j++)
			{
				cnt=0;
				if(!vis[i][j]&&ok(i,j)) 
				{
					dfs(i,j);	
		
					if(cnt>maxx)
					{
						maxx=cnt; 
					}
				}	
			}	
		} 
		cout<<maxx<<endl;
	}
	
}
#endif