Fox And Two Dots

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 4AAAAABCAAAAAOutputYesInput3 4AAAAABCAAADAOutputNoInput4 4YYYRBYBYBBBYBBBYOutputYesInput7 6AAAAABABBBABABAAABABABBBABAAABABBBABAAAAABOutputYesInput2 13ABCDEFGHIJKLMNOPQRSTUVWXYZOutputNo

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;}