Codeforces 510B Fox And Two Dots DFS

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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:

Codeforces 510B Fox And Two Dots DFS_第1张图片

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 - 1di 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).




解题思路:纯DFS。。刚开始发现如果搜回来会错。。于是做了一个记录数组。记录哪些点被访问过。

代码如下:

#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#define INF 0x0f0f0f0f
using namespace std;



int x,y;
char ma[55][55];
bool vis[55][55];
int flag=0;
bool dfs(char c,int i,int j,int step)
{
	if(flag)	return true;
	if(i==x&&j==y&&step>=4)
	{
		flag=1;
		return true;
	}
	if(vis[i][j]==1)	return false;
	else vis[i][j]=1;
	if(ma[i+1][j]==c) dfs(ma[i+1][j],i+1,j,step++);
	if(ma[i][j+1]==c) dfs(ma[i][j+1],i,j+1,step++);
	if(ma[i-1][j]==c) dfs(ma[i-1][j],i-1,j,step++);
	if(ma[i][j-1]==c) dfs(ma[i][j-1],i,j-1,step++);
	return false;
}

int main()
{
	int i,j,k,l,m,n;
	memset(ma,0,sizeof(ma));
	scanf("%d%d%*c",&m,&n);
	for(i=1;i<=m;i++)
	{
		for(j=1;j<=n;j++)
		{
			scanf("%c",&ma[i][j]);
		}
		scanf("%*c");
	}
/*
	for(i=1;i<=m;i++)
	{
		for(j=1;j<=n;j++)
		{

			printf("%c",ma[i][j]);
		}
		printf("\n");
	}*/
	for(i=1;i<=m;i++)
	{
		for(j=1;j<=n;j++)
		{	
			x=i;
			y=j;
			memset(vis,0,sizeof(vis));
			vis[i][j]=1;
			if(ma[i+1][j]==ma[i][j]) dfs(ma[i+1][j],i+1,j,1);
			if(ma[i][j+1]==ma[i][j]) dfs(ma[i][j+1],i,j+1,1);
			if(ma[i-1][j]==ma[i][j]) dfs(ma[i-1][j],i-1,j,1);
			if(ma[i][j-1]==ma[i][j]) dfs(ma[i][j-1],i,j-1,1);
			if(flag==1)
			{
				printf("Yes\n");
				return 0;
			}	
		}
	}
	if(flag==0)
	{
		printf("No\n");
	}
	return 0;

	
	
}



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