CoderForces 510B (dfs)

E - E Time Limit:2000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u Submit  Status  Practice  CodeForces 510B

Description

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, ..., d_k a cycle if and only if it meets the following condition:

These k dots are different: if i ≠ j then d_i is different from d_j.k is at least 4.All dots belong to the same color.For all 1 ≤ i ≤ k - 1: d_i and d_i + 1 are adjacent. Also, d_k 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.

Sample Input

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

Hint

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).

代码：

#include<stdio.h> #include<string.h> #include<algorithm> using namespace std; char map[51][51]; int vis[51][51]; int dx[4]={0,0,-1,1}; int dy[4]={1,-1,0,0}; char c; int st,en,n,m; bool flag; void dfs(int x,int y,int fx,int fy,char c)//fx,fy用来记录上一个能走的位置，避免向左走的时候重复； { if(flag==1) return ; for(int i=0;i<4;i++) { int a=x+dx[i]; int b=y+dy[i]; if(a>=0&&a<n&&b>=0&&b<m&&map[a][b]==c) { if(a==fx&&b==fy) { continue; } if(vis[a][b]==1)//走到起点才成立； { flag=1; return; } vis[a][b]=1;//走到一个能走的位置先标记再搜索； dfs(a,b,x,y,map[a][b]); } } } int main() { while(scanf("%d%d",&n,&m)!=EOF) { for(int i=0;i<n;i++) { scanf("%s",map[i]); } memset(vis,0,sizeof(vis)); flag=0; for(int i=0;i<n;i++) { for(int j=0;j<m;j++) { int fx=-1,fy=-1; if(vis[i][j]==0) { vis[i][j]=1; dfs(i,j,fx,fy,map[i][j]); if(flag==1) break; } } if(flag==1) break; } if(flag==1) printf("Yes\n"); else printf("No\n"); } return 0; }