codeforces B. Fox And Two Dots (dfs)

B. 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).

给一个n*m的map看有没有环

dfs遍历每个点，有环就跳出，最后要能回到起点，不能走刚刚遍历的点，所以dfs的参数处理一下

代码：

#include<cstdio> #include<cstring> #include<algorithm> using namespace std; char map[55][55]; int vis[55][55]; int dir[4][2]={{0,1},{0,-1},{1,0},{-1,0}}; int flag; int n,m; void dfs(int x,int y,int px,int py)//px,py是上一个点的位置， { if(vis[x][y]) { flag=1; return ; } vis[x][y]=1; for(int i=0;i<4;i++) { int nx=x+dir[i][0]; int ny=y+dir[i][1]; if(nx>=0&&nx<n&&ny>=0&&ny<m&&map[nx][ny]==map[x][y])//因为最后要回到起点，所以不做！vis[nx][ny]处理 { if(!(nx==px&&ny==py)) dfs(nx,ny,x,y); //避免下次再搜到 x,y } } return ; } int main() { while(~scanf("%d%d",&n,&m)) { for(int i=0;i<n;i++) scanf("%s",map[i]); for(int i=0;i<n;i++) { for(int j=0;j<m;j++) { memset(vis,0,sizeof(vis)); flag=0; dfs(i,j,-1,-1); if(flag==1) break; } if(flag==1) break; } if(flag==1) printf("Yes\n"); else printf("No\n"); } return 0; }