题目链接：http://codeforces.com/problemset/problem/510/B
B. Fox And Two Dots
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
Copy
3 4 AAAA ABCA AAAA
output
Copy
Yes
input
Copy
3 4 AAAA ABCA AADA
output
Copy
No
input
Copy
4 4 YYYR BYBY BBBY BBBY
output
Copy
Yes
input
Copy
7 6 AAAAAB ABBBAB ABAAAB ABABBB ABAAAB ABBBAB AAAAAB
output
Copy
Yes
input
Copy
2 13 ABCDEFGHIJKLM NOPQRSTUVWXYZ
output
Copy
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：给出一个迷宫，求能不能找到一个环；
那次比赛差一丢丢丢丢丢丢.......（还是太菜）
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int n,m;
char vec[55][55];
int vis[55][55];
int flag=0;
int dir[4][2]={{1,0},{-1,0},{0,1},{0,-1}};
void dfs(int bx,int by,int fx,int fy,char ch){
if(flag){
return ;
}
for(int i=0;i<4;i++){
int x= bx + dir[i][0];
int y= by + dir[i][1];
if(x>=0 && x<n && y>=0 && y<m && vec[x][y]==ch){
if(x==fx && y==fy){
continue;
}
if(vis[x][y]){
flag=1;
return ;
}else{
vis[x][y]=1;
dfs(x,y,bx,by,ch);
}
}
}
}
int main(){
cin>>n>>m;
memset(vis,0,sizeof(vis));
for(int i=0;i<n;i++){
scanf("%s",vec[i]);
}
for(int i=0;i<n;i++){
for(int j=0;j<m;j++){
if(vis[i][j]!=1){
vis[i][j]=1;
flag=0;
dfs(i,j,-5,-5,vec[i][j]);
if(flag){
break;
}
}
}
if(flag){
break;
}
}
if(flag){
cout<<"Yes"<<endl;
}
else{
cout<<"No"<<endl;
}
return 0;
}