Fox And Two Dots（DFS）

最新推荐文章于 2022-02-01 01:07:42 发布

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最新推荐文章于 2022-02-01 01:07:42 发布

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这是一篇关于解决移动游戏“Two Dots”中寻找相同颜色环形路径问题的博客。通过输入的二维矩阵表示游戏板，题目要求确定是否存在同色的环形路径。通过DFS算法，并使用标记vis来判断在遍历过程中是否形成环，当步数大于4时，说明存在环形路径。给出了一些样例测试用例并解释了判断条件。

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Fox And Two Dots

Fox Ciel is playing a mobile puzzle game called “Two Dots”. The basic levels are played on a board of size n × m cells, like this:

Each cell contains a dot that has some color. We will use different uppercase Latin characters to express different colors.

The key of this game is to find a cycle that contain dots of same color. Consider 4 blue dots on the picture forming a circle as an example. Formally, we call a sequence of dots d1, d2, …, dk a cycle if and only if it meets the following condition:

These k dots are different: if i ≠ j then di is different from dj.
k is at least 4.
All dots belong to the same color.
For all 1 ≤ i ≤ k - 1: di and di + 1 are adjacent. Also, dk and d1 should also be adjacent. Cells x and y are called adjacent if they share an edge.
Determine if there exists a cycle on the field.

Input
The first line contains two integers n and m (2 ≤ n, m ≤ 50): the number of rows and columns of the board.

Then n lines follow, each line contains a string consisting of m characters, expressing colors of dots in each line. Each character is an uppercase Latin letter.

Output
Output “Yes” if there exists a cycle, and “No” otherwise.

Examples
Input
3 4
AAAA
ABCA
AAAA
Output
Yes
Input
3 4
AAAA
ABCA
AADA
Output
No
Input
4 4
YYYR
BYBY
BBBY
BBBY
Output
Yes
Input
7 6
AAAAAB
ABBBAB
ABAAAB
ABABBB
ABAAAB
ABBBAB
AAAAAB
Output
Yes
Input
2 13
ABCDEFGHIJKLM
NOPQRSTUVWXYZ
Output
No
Note
In first sample test all ‘A’ form a cycle.

In second sample there is no such cycle.

The third sample is displayed on the picture above (‘Y’ = Yellow, ‘B’ = Blue, ‘R’ = Red).
这道题是判断相同颜色（字母）是否成环，两种方法：1：由相同颜色dfs中第一次遇到标记过的vis判断step是否大于4，大于4即成环。2：判断遍历到的颜色不是上一个坐标，即成环
第二种方法

#include<iostream>
#include<algorithm>
#include<queue>
#include<cstring>
using namespace std;

int d[4][2]={0,1,1,0,0,-1,-1,0};//上，右，下，左 
int vis[55][55];//标记,当遍历到已经标记过的非上一个点，即成环 
char s[55][55];//存图
int flag;
int m,n;
char k;
int ans=0;
int zm[26]={0};


void dfs(int x,int y,int prex,int prey){
	if(vis[x][y]){
				flag=1;
				return;
			}
	vis[x][y]=1;
	for(int i=0;i<4;i++){
		int xx=x+d[i][0];
		int yy=y+d[i][1];
		if(xx>=0&&xx<n&&yy>=0&&yy<m&&s[xx][yy]==s[prex][prey]){
		//	cout<<xx<<"&&&"<<yy<<" "<<prex<<"%%%"<<prey<<endl;
			if(xx==prex&&yy==prey){
			//	cout<<xx<<"chongfu"<<yy<<endl;
				continue;
			}
	
			
			
				dfs(xx,yy,x,y);
			//	vis[xx][yy]=0;
		
			
			
		}
	}
	
	
}

int main(){
	cin>>n>>m;//n为行数，m为列数 
	flag=0;
		
		for(int i=0;i<n;i++){
			for(int j=0;j<m;j++){
				cin>>s[i][j];
				
			}
		}
			memset(vis,0,sizeof(vis));
		for(int i=0;i<n;i++){
			for(int j=0;j<m;j++){
			
				
				if(!vis[i][j]){
				
						dfs(i,j,i,j);
				
				}
				
				if(flag){
					//	cout<<"2222"<<endl;
						break;
					}
			}
			if(flag){
					//	cout<<"3333"<<endl;
						break;
					}
		}
		if(flag)
		cout<<"Yes"<<endl;
		else cout<<"No"<<endl;
	
	return 0;
}