B. Fox And Two Dots Codeforces Round #290 (Div. 2)

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:

1. These k dots are different: if i ≠ j then di is different from dj.
2. k is at least 4.
3. All dots belong to the same color.
4. 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 4AAAAABCAAAAA

output

Yes

input

3 4AAAAABCAAADA

output

No

input

4 4YYYRBYBYBBBYBBBY

output

Yes

input

7 6AAAAABABBBABABAAABABABBBABAAABABBBABAAAAAB

output

Yes

input

2 13ABCDEFGHIJKLMNOPQRSTUVWXYZ

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).

题意：在n*m的地图上，每个格子涂有不同的颜色（A~Z），问是否有某一种颜色能组成一个回路且长度至少为4.

思路：我用的bfs，从一个未被访问的点开始bfs，走同一种颜色的格子，同时不能走回头路，如果存在两条路最后能碰头并且长度之和大于4就输出Yes，否则No。

代码：

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#include <string>#include <map>#include <stack>#include <vector>#include <set>#include <queue>#define maxn 55using namespace std;struct Node{    int x,y,step,d;};int dir[4][2]={-1,0,0,1,1,0,0,-1};char mp[maxn][maxn];int step[maxn][maxn];bool vis[maxn][maxn];int n,m;bool isok(int x,int y){    if (x>=0&&x<n&&y>=0&&y<m)        return true;    return false;}bool bfs(int x,int y){    Node st,now;    queue<Node>Q;    while (!Q.empty())        Q.pop();    st.x=x;st.y=y;st.step=0;st.d=-1;    Q.push(st);    step[x][y]=0;    vis[x][y]=true;    while (!Q.empty())    {        st=Q.front(); Q.pop();        for (int i=0;i<4;i++)        {            if (st.d==(i+2)%4) continue;   //不能走回头路            now.x=st.x+dir[i][0];            now.y=st.y+dir[i][1];            now.step=st.step+1;            now.d=i;    //记录当前点走到下一个点是从哪个方向过去的            if (isok(now.x,now.y)&&mp[st.x][st.y]==mp[now.x][now.y])            {                if (vis[now.x][now.y]&&(now.step+step[now.x][now.y]+1)>=4)  //碰头了并且长度之和大于4                    return true;                if (!vis[now.x][now.y])  //没有被访问过就入队列                {                    vis[now.x][now.y]=true;                    step[now.x][now.y]=now.step;                    Q.push(now);                }            }        }    }    return false;}bool solve(){    for (int i=0;i<n;i++)        for (int j=0;j<m;j++)            if (!vis[i][j]&&bfs(i,j))                return true;    return false;}int main(){    while (~scanf("%d%d",&n,&m))    {        for (int i=0;i<n;i++)            scanf("%s",mp[i]);        memset(vis,false,sizeof(vis));        memset(step,0,sizeof(step));        if (solve()) printf("Yes\n");        else printf("No\n");    }    return 0;}