B. Fox And Two Dots

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 test(s)

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).

法一：bfs 这题一开始把队列开成10^7，结果超时了，后来再把队列开成10^5,15ms过了，这里注意几点：开方向数组的时候，每一个要用大括号，不能用小括号，还有写bfs的时候，一开始front=rear=0;之后每一个数据进队，rear++,再是q[rear].x=x;q[rear].y=y;判断退出的条件是while(front<=rear)

#include<stdio.h>

#include<string.h>

char a[200][200];

int m,n,flag,t;

int tab[10][2]={{0,0},{0,1},{-1,0},{0,-1},{1,0}},vis[200][200];

struct node

{

int x,y;

}q[11111],pre[11111];

void bfs(int x,int y)

{

int i,j,front=0,rear=0,xx,yy,x1,y1;

memset(q,-1,sizeof(q));

q[rear].x=x;

q[rear].y=y;

memset(pre,-1,sizeof(pre));

while(front<=rear)

{

x1=q[front].x;

y1=q[front].y;

vis[x1][y1]=1;

front++;

for(i=1;i<=4;i++)

{

xx=x1+tab[i][0];

yy=y1+tab[i][1];

if(xx>=0 && xx<n && yy>=0 && yy<m && a[xx][yy]==a[x1][y1]){

if(pre[front-1].x==xx && pre[front-1].y==yy)

continue;

if(vis[xx][yy]==1){

flag=1;break;

}

rear++;vis[xx][yy]=1;

q[rear].x=xx;q[rear].y=yy;pre[rear].x=x1;pre[rear].y=y1;

}

}

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",a[i]);

memset(vis,0,sizeof(vis));

for(i=0;i<n;i++){

for(j=0;j<m;j++){

if(vis[i][j]==1)

continue;

flag=0;

bfs(i,j);

if(flag==1)

break;

}

if(flag==1)break;

}

if(flag==1){

printf("Yes\n");

}

else printf("No\n");

}

return 0;

}

法二：dfs 要注意不能走回头路

#include<stdio.h>

#include<string.h>

char a[200][200];

int m,n,flag,t;

int tab[10][2]={{0,0},{0,1},{-1,0},{0,-1},{1,0}},vis[200][200];

void dfs(int x,int y,int prex,int prey)

{

int i,j,xx,yy;

for(i=1;i<=4;i++){

xx=x+tab[i][0];

yy=y+tab[i][1];

if(xx>=0 && xx<n && yy>=0 && yy<m && a[xx][yy]==a[x][y]){

if(xx==prex && yy==prey){

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",a[i]);

memset(vis,0,sizeof(vis));

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==1){

printf("Yes\n");

}

else printf("No\n");

}

return 0;

}