【Codeforces】-510B-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).

题意：就是判断相同字母可不可以形成一个环。最少4个才可以形成回路，其实不需要考虑步数大不大于4，形成了回路一定不小于4.

题解：标记每个走过的点，下一个遇到的点不是传给他的点而且被标记就说明有回路。

#include<cstdio> #include<cstring> #include<algorithm> using namespace std; char a[55][55]; int vis[55][55]; int dx[4]={0,0,1,-1}; int dy[4]={1,-1,0,0}; int n,m,ans; void dfs(int x,int y,int fx,int fy ) //记录自己和传给自己的点的坐标 { if(vis[x][y]) { ans=1; return ; //遇到走过的点表示形成了回路，就是可以，返回 } vis[x][y]=1; int nx,ny; for(int i=0;i<4;i++) { nx=x+dx[i]; ny=y+dy[i]; if(nx==fx&&ny==fy) //遇到上一个点，继续 continue; if(a[nx][ny]==a[x][y]) //和上一个点相同，对这个点查找 dfs(nx,ny,x,y); } } int main() { while(~scanf("%d %d",&n,&m)) { ans=0; memset(vis,0,sizeof(vis)); for(int i=1;i<=n;i++) scanf("%s",a[i]+1); for(int i=1;i<=n;i++) { for(int j=1;j<=m;j++) { if(!vis[i][j]) //遍历没有走过的点 dfs(i,j,i,j); if(ans) break; } if(ans) break; } if(ans) printf("Yes\n"); else printf("No\n"); } }