CodeForces 510B Fox And Two Dots(深搜DFS)

题目：

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).

思路：

跟走迷宫差不多，注意标记上一步的，最后要走回来

代码:

#include <cstdio> #include <cstring> #include <cstdlib> #include <cmath> #include <string> #include <iostream> #include <stack> #include <queue> #include <vector> #include <algorithm> #define mem(a,b) memset(a,b,sizeof(a)) #define N 600+20 #define M 200010 #define MOD 1000000000+7 #define inf 0x3f3f3f3f using namespace std; char map[200][200]; int m,n,flag,t; int go[4][2]= {{0,1},{-1,0},{0,-1},{1,0}},vis[200][200]; void dfs(int x,int y,int fx,int fy) { int i,j,xx,yy; for(i=0; i<4; i++) { xx=x+go[i][0]; yy=y+go[i][1]; if(xx>=0&&xx<n&&yy>=0 && yy<m && map[xx][yy]==map[x][y]) { if(xx==fx && yy==fy) { continue; } if(vis[xx][yy]==1) { flag=1; break; } vis[xx][yy]=1; dfs(xx,yy,x,y); } if(flag==1) break; } return; } int main() { int i,j,u,v; while(scanf("%d%d",&n,&m)!=EOF) { for(i=0; i<n; i++) scanf("%s",map[i]); mem(vis,0); for(i=0; i<n; i++) { for(j=0; j<m; j++) { if(vis[i][j]==1) continue; flag=0; vis[i][j]=1; dfs(i,j,-1,-1); if(flag==1) break; } if(flag==1) break; } if(flag) printf("Yes\n"); else printf("No\n"); } return 0; }