CodeForces 510B Fox And Two Dots （DFS）

寻找字母环游戏算法

最新推荐文章于 2020-11-25 13:52:32 发布

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最新推荐文章于 2020-11-25 13:52:32 发布

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分类专栏： DFS

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本文介绍了一款名为“Two Dots”的拼图游戏中的算法挑战，玩家需在棋盘上找出相同颜色的字母形成的环。文章详细解释了游戏规则，并通过示例说明如何判断是否存在符合条件的字母环。

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
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).

只能说自己对DFS不够理解吧，才致使不能灵活运用

题意：判断是否有字母成环。

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int dx[]={0,0,1,-1};
int dy[]={-1,1,0,0};
char map[55][55];
int vis[55][55];
int n,m;
bool dfs(int x,int y,int sx,int sy,char c)
{
	vis[x][y]=1;
	for(int i=0;i<4;i++)
	{
		int tx=x+dx[i];
		int ty=y+dy[i];
		if(tx==sx&&ty==sy)
		continue;
		if(tx>=0&&ty>=0&&tx<n&&ty<m&&map[tx][ty]==c)
		{
			if(vis[tx][ty]==1)
			return 1;
			if(dfs(tx,ty,x,y,c))
			return 1;
		}
		
	}
	return 0;
}
int main()
{
	int i,j;
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		for(i=0;i<n;i++)
		scanf("%s",map[i]);
		for(i=0;i<n;i++)
		for(j=0;j<m;j++)
		{
			if(!vis[i][j])
			{
				if(dfs(i,j,-1,-1,map[i][j]))
				{
					printf("Yes\n");
					return 0;
				}
			}
			
		}
		printf("No\n");
	}
	return 0;
}