Fox And Two Dots
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 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.
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).
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<time.h>
#include<iostream>
#include<algorithm>
#include<string>
#include<vector>
#include<queue>
#include<set>
#include<map>
#include<stack>
using namespace std;
#define LL long long
#define mst(a) memset(a,0,sizeof(a))
const int max_n=2e5+5;
int vis[51][51],a[51][51],n,m,flag;
int ii,jj;
char s[51];
int mynext[4][2]={{0,1},{1,0},{-1,0},{0,-1}};
void dfs(int x,int y,int step)
{
for(int k=0;k<=3;k++)
{
int next_x=x+mynext[k][0],next_y=y+mynext[k][1];
if(next_x>=n||next_y>=m||next_y<0||next_x<0)continue;
if(x==ii&&jj==y&&step>2)
{
flag=1;
return ;
}
if(a[next_x][next_y]==a[x][y]&&vis[next_x][next_y]==0)
{
vis[next_x][next_y]=1;
dfs(next_x,next_y,step+1);
}
}
return ;
}
int main()
{
//freopen("1.in","r",stdin);freopen("1.out","w",stdout);
scanf("%d%d",&n,&m);
for(int i=0;i<n;i++)
{
scanf("%s",s);
for(int j=0;j<m;j++)
{
a[i][j]=s[j]-'A';
}
}
for(int i=0;i<n;i++)
{
for(int j=0;j<m;j++)
{
mst(vis);
ii=i,jj=j;
dfs(i,j,0);
}
}
if(flag)printf("Yes\n");
else printf("No\n");
return 0;
}