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).

只能说自己对DFS不够理解吧，才致使不能灵活运用 题意：判断是否有字母成环。 #include<stdio.h> #include<string.h> #include<algorithm> using namespace std; int dx[]={0,0,1,-1}; int dy[]={-1,1,0,0}; char map[55][55]; int vis[55][55]; int n,m; bool dfs(int x,int y,int sx,int sy,char c) { vis[x][y]=1; for(int i=0;i<4;i++) { int tx=x+dx[i]; int ty=y+dy[i]; if(tx==sx&&ty==sy) continue; if(tx>=0&&ty>=0&&tx<n&&ty<m&&map[tx][ty]==c) { if(vis[tx][ty]==1) return 1; if(dfs(tx,ty,x,y,c)) return 1; } } return 0; } int main() { int i,j; while(scanf("%d%d",&n,&m)!=EOF) { for(i=0;i<n;i++) scanf("%s",map[i]); for(i=0;i<n;i++) for(j=0;j<m;j++) { if(!vis[i][j]) { if(dfs(i,j,-1,-1,map[i][j])) { printf("Yes\n"); return 0; } } } printf("No\n"); } return 0; }