[CodeForces510B]

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:

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).

题为同一种字母是否能够围成一个圈

#include<cstdio>#include<cstring>#include<algorithm>#include<queue>using namespace std;#include<cmath>int m,n,ans,ex,ey,ok;const int MAX=55;typedef long long LL;char s[MAX][MAX];int vis[MAX][MAX];int fx[4]={0,0,1,-1};int fy[4]={1,-1,0,0};void dfs(int x,int y,int nl){    if(ok) return ;   //成环结束    vis[x][y]=1;    for(int i=0;i<4;i++)    {        int xx=fx[i]+x;        int yy=fy[i]+y;        if(xx==ex&&yy==ey&&nl>2){   //至少四个字母            ok=1;            return ;        }        if(xx>=0&&yy>=0&&xx<m&&yy<n&&!vis[xx][yy]&&s[xx][yy]==s[x][y])        dfs(xx,yy,nl+1);    }}int main(){    scanf("%d%d",&m,&n);    for(int i=0;i<m;i++)    scanf("%s",&s[i]);    ok=0;    for(int i=0;i<m&&!ok;i++)    {        for(int j=0;j<n&&!ok;j++)        {            memset(vis,0,sizeof(vis));  //初始化为0            ex=i;ey=j;            dfs(i,j,0);        }    }    if(!ok) printf("No");    else printf("Yes");return 0;}