Description
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.
Sample Input
3 4 AAAA ABCA AAAA
Yes
3 4 AAAA ABCA AADA
No
4 4 YYYR BYBY BBBY BBBY
Yes
7 6 AAAAAB ABBBAB ABAAAB ABABBB ABAAAB ABBBAB AAAAAB
Yes
2 13 ABCDEFGHIJKLM NOPQRSTUVWXYZ
No
Hint
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).
简单来说就是找一个环 这个环由相同的字母构成 用dfs从每个点出发看能不能再回到初始点
为了防止 出现下面的情况加入了fx fy
AAA
ABA
AAA
(0,0)—>(0,1)—>(0,0)这样没有形成环就直接结束了 加入fx fy是为了防止回头查找 也就是只能让查找不能回头
#include<cstdio>
#include<cstring>
using namespace std;
char map[101][101];
int vis[101][101];
int bg,flog;
int n,m;
int dx[4]={0,-1,1,0};
int dy[4]={-1,0,0,1};
void dfs(int x,int y,int fx,int fy,char s)
{
if(flog==1)
{
return ;
}
for(int i=0;i<4;i++)
{
int nx=x+dx[i];
int ny=y+dy[i];
if(nx>=0&&ny>=0&&nx<n&&ny<m&&map[nx][ny]==s)
{
if(nx==fx&&ny==fy)
{
continue;
}
if(vis[nx][ny]==1&&map[nx][ny]==s)
{
flog=1;
return ;
}
vis[nx][ny]=1;
dfs(nx,ny,x,y,s);
}
}
}
int main()
{
memset(vis,0,sizeof(vis));
scanf("%d%d",&n,&m);
for(int i=0;i<n;i++)
{
scanf("%s",map[i]);
}
flog=0;
for(int i=0;i<n;i++)
{
for(int j=0;j<m;j++)
{
if(vis[i][j]==0)
{
vis[i][j]=1;
dfs(i,j,-2,-2,map[i][j]);
if(flog==1)
{
break;
}
}
}
if(flog==1)
{
break;
}
}
if(flog==1)
{
printf("Yes
");
}
else
{
printf("No
");
}
return 0;
}