codeforces#290 B&&510 B Fox And Two Dots(简单dfs)

B. Fox And Two Dots

点击打开题目链接

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Fox Ciel is playing a mobile puzzle game called "Two Dots". The basic levels are played on a board of sizen × 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 dotsd₁, d₂, ..., d_k acycle if and only if it meets the following condition:

1. These k dots are different: if i ≠ j then d_i is different fromd_j.
2. k is at least 4.
3. All dots belong to the same color.
4. For all 1 ≤ i ≤ k - 1: d_i and d_i + 1 are adjacent. Also,d_k andd₁ should also be adjacent. Cellsx 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 andm (2 ≤ n, m ≤ 50): the number of rows and columns of the board.

Then n lines follow, each line contains a string consisting ofm 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.

Sample test(s)

Input

3 4AAAAABCAAAAA

Output

Yes

Input

3 4AAAAABCAAADA

Output

No

Input

4 4YYYRBYBYBBBYBBBY

Output

Yes

Input

7 6AAAAABABBBABABAAABABABBBABAAABABBBABAAAAAB

Output

Yes

Input

2 13ABCDEFGHIJKLMNOPQRSTUVWXYZ

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：判断有没有环，元素是4个以上的，简单dfs；

代码：

#include <iostream>#include<stdio.h>#include<string.h>#include<stdlib.h>#include<vector>#include<stdlib.h>#include<string>#include<map>#include<set>#include<queue>#include<stack>#include<math.h>#define inf 0x3f3f3f3f#define eps 1e-5#define max(a,b) (a)>(b)?(a):(b)#define min(a,b) (a)<(b)?(a):(b)using namespace std;char map1[51][51];bool visit[51][51];int dir[][2]={{1,0},{-1,0},{0,1},{0,-1}};int n,m,step;bool dfs(char c,int sx,int sy,int cx,int cy){    if(sx==cx&&sy==cy&&visit[cx][cy])    {        return step>=4?true:false;    }    int i;    visit[cx][cy]=1;    for(i=0;i<4;i++)    {        int nx=cx+dir[i][0];        int ny=cy+dir[i][1];        if((nx>=0&&nx<n&&ny>=0&&ny<m&&map1[nx][ny]==c&&!visit[nx][ny])||(nx==sx&&ny==sy))        {            step++;            if(dfs(c,sx,sy,nx,ny))                return true;            step--;        }    }    return false;}bool slove(){    int i,j;    for(i=0;i<n;i++)    {        for(j=0;j<m;j++)        {            memset(visit,0,sizeof(visit));            step=0;            if(dfs(map1[i][j],i,j,i,j))                return true;        }    }    return false;}int main(){    int i;    while(~scanf("%d%d",&n,&m))    {        for(i=0; i<n; i++)            scanf("%s",map1[i]);        bool ans=slove();        printf("%s\n",ans?"Yes":"No");    }    return 0;}