codeforces#290 B&&510 B Fox And Two Dots (简单dfs)

最新推荐文章于 2020-09-12 18:46:10 发布

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本文介绍了一个基于游戏“Two Dots”的算法挑战，目标是在给定的矩阵中寻找由相同颜色圆点构成的循环。通过深度优先搜索（DFS）实现解决方案，并提供了一段AC代码作为示例。

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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:

$\text{[math]}$

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
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
ABABBB
ABAAAB
ABAAAB
AAAAAB
ABBBAB

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).

DFS还是依旧写不出来。。。。。。。

AC代码：

#include <cstdio>  
#include <cstring>  
#include <iostream>  
#include <algorithm>  
using namespace std;  
char s[56][56];  
int vis[56][56];  
int n, m;  
int xx[4] = {0,-1,1,0};  
int yy[4] = {1,0,0,-1};  
char tt;  
int mark;  
int judge(int x, int y)  
{  
    if(x>=0 && x<n && y>=0 && y<m)  
        return 1;  
    else  
        return 0;  
}  
  
void dfs(int x, int y, int fx, int fy)  
{  
    if(!judge(x, y))  
        return ;  
    vis[x][y] = 1;  
    for(int i = 0; i < 4; i++)  
    {  
        int tx = x + xx[i];  
        int ty = y + yy[i];  
        if(judge(tx,ty) && s[x][y]==s[tx][ty] && (tx!=fx || ty!=fy))  
        {  
            if(vis[tx][ty])  
            {  
                mark = 1;  
                return ;  
            }  
            dfs(tx, ty, x, y);  
        }  
    }  
    return ;  
}  
  
int main()  
{  
    while(~scanf("%d%d",&n,&m))  
    {  
        memset(vis,0,sizeof(vis));  
        for(int i = 0; i < n ; i++)  
        {  
            scanf("%s",s[i]);  
        }  
        mark = 0;  
        for(int i = 0; i < n; i++)  
        {  
            for(int j = 0; j < m; j++)  
            {  
                if(!vis[i][j])  
                {  
                    dfs(i,j,-1,-1);  
                    if(mark)  
                    {  
                        printf("Yes\n");  
                        return 0;  
                    }  
                }  
            }  
        }  
        printf("No\n");  
    }   
}