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 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：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 55
using 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;
}

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