Codeforces_Round_290_(Div._2)_B._Fox_And_Two_Dots

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题目链接 : http://codeforces.com/contest/510/problem/B

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 d 1 , d 2 , ..., d k a cycle if and only if it meets the following condition:

1. These k dots are different: if i ≠ j then d i is different from d j .
2. k is at least 4.
3. All dots belong to the same color.
4. For all 1 ≤ i ≤ k - 1 : d i and d i + 1 are adjacent. Also, d k and d 1 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

DFS 判断是否存在环 。

#include<stdio.h>
#include<iostream>
#include<math.h>
#include<stdlib.h>
#include<ctype.h>
#include<algorithm>
#include<vector>
#include<string>
#include<queue>
#include<stack>
#include<set>
#include<map>
#include<string>
#include<sstream>

using namespace std;

int n, m, ok;
char s[100][100];
int vis[100][100];
int dir[4][2] = { { 1, 0 }, { -1, 0 }, { 0, 1 }, { 0, -1 } };
int is_ok(int x, int y)
{
	if (x >= 0 && x < n && y >= 0 && y < m)
		return 1;
	return 0;
}

int dfs(int x, int y, int prex, int prey, char c)
{
	vis[x][y] = 1;
	for (int i = 0; i < 4; i++)
	{
		int xx = x + dir[i][0];
		int yy = y + dir[i][1];
		if (xx == prex && yy == prey) continue;
		if (is_ok(xx, yy) && s[xx][yy] == c)
		{
			if (vis[xx][yy])
				return 1;
			if (dfs(xx, yy, x, y, c))
				return 1;
		}
	}
	return 0;
}

int main()
{
	while (cin >> n >> m)
	{
		ok = 0; memset(vis, 0, sizeof(vis));
		for (int i = 0; i < n; i++)
			cin >> s[i];

		for (int i = 0; i < n; i++)
			for (int j = 0; j < m; j++)
			{
				if (!vis[i][j])
					if (dfs(i, j, -1, -1, s[i][j]))
					{
						cout << "Yes" << endl;
						ok = 1;
						return 0;
					}
			}
		if (!ok)
			cout << "No" << endl;
	}
	return 0;
}