codeforces 510b Fox And Two Dots

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).

题意：给定一个图，找环（有相同的颜色）

代码：

#include<cstdio> #include<cstring> #include<algorithm> using namespace std; const int maxn=55; char maze[maxn][maxn]; int vis[maxn][maxn]; int dx[4]= {0,1,0,-1}; int dy[4]= {1,0,-1,0}; int N,M,sx,sy; int cnt=0,ok=0; void dfs(int x,int y,char p) { for(int i=0; i<4; i++) { int nx=x+dx[i],ny=y+dy[i]; if(nx>=0&&nx<N&&ny>=0&&ny<M&&maze[x][y]==p) { if(!vis[nx][ny]) { vis[nx][ny]=1; //修改全局变量； cnt++; dfs(nx,ny,p); cnt--; vis[nx][ny]=0; //切记一定要修改过来！ } if(nx==sx&&ny==sy&&(cnt>2)) //表示形成了环！ { ok=1; break; } } if(ok) break; } return ; } int main() { while(~scanf("%d%d",&N,&M)) { for(int i=0; i<N; i++) scanf("%s",maze[i]); ok=0; for(int i=0; i<N; i++) { for(int j=0; j<M; j++) { memset(vis,0,sizeof(vis)); vis[i][j]=1; cnt=0; sx=i,sy=j; dfs(i,j,maze[i][j]); if(ok) break; } if(ok) break; } if(ok) printf("Yes\n"); else printf("No\n"); } return 0; }