Fox And Two Dots

最新推荐文章于 2022-02-01 01:07:42 发布

最新推荐文章于 2022-02-01 01:07:42 发布 · 235 阅读

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本文介绍了一种使用深度优先搜索（DFS）算法来解决TwoDots游戏中寻找相同颜色圆点形成的有效循环的问题。通过避免回头路径确保了搜索的有效性，并提供了一个具体的实现示例。

题目来源于2017HPUACM暑期培训：https://vjudge.net/contest/174968#problem/D

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.

Example
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).

也是dfs的运用，但是注意这里面dfs不能回头，加变量判断它不走重复路线

#include<cstdio>
#include<cmath>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<iostream>
#include<queue>
using namespace std;
int n,m,k=0,vis[55][55],xx,yy;
char s[55][55];
int fx[4]={1,0,0,-1};
int fy[4]={0,-1,1,0};
void dfs(int x,int y,int px,int py,char ss)//px，py判断是否回头 
{
    if(k==1)
    {
        return;
    }
    for(int i=0;i<4;i++)
    {
        xx=x+fx[i];
        yy=y+fy[i];
        if(xx>=0&&yy>=0&&xx<n&&yy<m&&s[xx][yy]==ss)
        {
            if(xx==px&&yy==py)//px，py判断是否回头
            {
                continue;
            }
            else
            {
                if(vis[xx][yy]&&s[xx][yy]==ss)
                {
                    k=1;
                    return ;
                }
            }
            vis[xx][yy]=1;
            dfs(xx,yy,x,y,s[xx][yy]);
        }
    }
}
int main()
{
    k=0;
    memset(vis,0,sizeof(vis));
    scanf("%d%d",&n,&m);
    for(int i=0;i<n;i++)
       scanf("%s",s[i]);
    for(int i=0;i<n;i++)
    {
        for(int j=0;j<m;j++)
        {
            if(!vis[i][j])
            {
                vis[i][j]=1;
                dfs(i,j,-1,-1,s[i][j]);
                if(k==1)
                  break;
            }
        }
        if(k==1)
          break;
    }
    if(k==1)
      printf("Yes\n");
    else
      printf("No\n");
return 0;
}