B. Fox And Two Dots

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Manton过

最新推荐文章于 2024-12-27 17:06:57 发布

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本文介绍了一款名为“Two Dots”的游戏中的算法挑战，任务是在一个由不同颜色圆点组成的网格中找到由相同颜色圆点形成的矩形。通过深度优先搜索（DFS）的方法来判断是否存在这样的矩形。

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点击打开题目

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:

$\text{[math]}$

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).

题意：利用不同字母表示不同颜色，问是否存在由同种颜色组成的矩形。

题解：利用dfs 搜索，由于dfs只能向上下左右四个方向走，如果能回到起点，则是矩形。

代码：

#include<iostream>
#include<cstring>
#define N 55
using namespace std;

int n,m,flag,ex,ey;
int vis[N][N];
char mapp[N][N];
int fx[4]={0,0,1,-1},fy[4]={1,-1,0,0};

void dfs(int x,int y,int nl){
	if(flag)return ;
	vis[x][y]=1;
	for(int i=0;i<4;i++){
		int nx=x+fx[i];
		int ny=y+fy[i];
		if(nx==ex&&ny==ey&&nl>2){//判断是否回到起点，且步数大于2,（防止AA这种情况出现）
			flag=1;
			return ;
		}
		if(nx >= 0 && nx < n && ny >= 0 && ny < m && !vis[nx][ny] && mapp[nx][ny]==mapp[x][y])
		dfs(nx,ny,nl+1);
	}
}

int main(){
	while(cin>>n>>m){
		for(int i=0;i<n;i++)
		cin>>mapp[i];
		flag=0;
		for(int i=0;i<n&&!flag;i++){
			for(int j=0;j<m && !flag;j++){
				memset(vis,0,sizeof(vis));
				ex=i;ey=j;
				dfs(i,j,0);
			}
		}
		if(flag)
		cout<<"Yes"<<endl;
		else
		cout<<"No"<<endl;	
	}
	return 0;
}