codeforces#290 B&&510 B Fox And Two Dots (简单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 ABABBB ABAAAB ABAAAB AAAAAB ABBBAB 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).

DFS还是依旧写不出来。。。。。。。

AC代码：

#include <cstdio> #include <cstring> #include <iostream> #include <algorithm> using namespace std; char s[56][56]; int vis[56][56]; int n, m; int xx[4] = {0,-1,1,0}; int yy[4] = {1,0,0,-1}; char tt; int mark; int judge(int x, int y) { if(x>=0 && x<n && y>=0 && y<m) return 1; else return 0; } void dfs(int x, int y, int fx, int fy) { if(!judge(x, y)) return ; vis[x][y] = 1; for(int i = 0; i < 4; i++) { int tx = x + xx[i]; int ty = y + yy[i]; if(judge(tx,ty) && s[x][y]==s[tx][ty] && (tx!=fx || ty!=fy)) { if(vis[tx][ty]) { mark = 1; return ; } dfs(tx, ty, x, y); } } return ; } int main() { while(~scanf("%d%d",&n,&m)) { memset(vis,0,sizeof(vis)); for(int i = 0; i < n ; i++) { scanf("%s",s[i]); } mark = 0; for(int i = 0; i < n; i++) { for(int j = 0; j < m; j++) { if(!vis[i][j]) { dfs(i,j,-1,-1); if(mark) { printf("Yes\n"); return 0; } } } } printf("No\n"); } }