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

梦里梦外; 2022-09-25 05:18 102阅读 0赞

E - E

**Time Limit:**2000MS **Memory Limit:**262144KB **64bit IO Format:**%I64d & %I64u

Submit [Status][] [Practice][] [CodeForces 510B][]

Description

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:

![Center][]

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 Input

Input

3 4
    AAAA
    ABCA
    AAAA

Output

Yes

Input

3 4
    AAAA
    ABCA
    AADA

Output

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

Hint

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).

我用深度优先搜索做的，

ac代码：

#include<cstdio>
    #include<iostream>
    #include<algorithm>
    #include<cstring>
    using namespace std;
    const int N = 55;
    int px[4]= {1,-1,0,0};
    int py[4]= {0,0,1,-1};
    char map[N][N];
    bool vis[N][N];
    int temp,c1,c2,n,m;
    bool check(int x,int y) {
    	int flag=0;
    	for(int k=0; k<4; k++) {
    		int u=x+px[k];
    		int v=y+py[k];
    		if(u>=0&&v>=0&&u<n&&v<m&&map[u][v]==map[c1][c2]&&vis[u][v])
    			flag++;
    	}
    //	printf("%d---\n",flag);
    	if(flag>=2)
    		return true;
    	return false;
    }
    bool judge(int x,int y) {
    	if(x>=0&&y>=0&&x<n&&y<m&&map[x][y]==map[c1][c2]&&!vis[x][y])
    		return true;
    	return false;
    }
    void dfs(int i,int j,int cnt) {
    	cnt++;
    	vis[i][j]=true;
    	if(check(i,j)&&cnt>=4)
    		temp=1;
    	for(int l=0; l<4; l++) {
    		int nx=i+px[l];
    		int ny=j+py[l];
    		if(judge(nx,ny)) {
    			vis[nx][ny]=true;
    			dfs(nx,ny,cnt);
    		}
    	}
    }
    int main() {
    	while(scanf("%d%d",&n,&m)!=EOF) {
    		temp=0;
    		for(int i=0; i<n; i++) {
    			scanf("%s",map[i]);
    		}
    		memset(vis,false,sizeof(vis));
    		for(int i=0; i<n; i++) {
    			for(int j=0; j<m; j++) {
    				if(!vis[i][j]) {
    					c1=i,c2=j;
    					dfs(i,j,0);
    				}
    			}
    		}
    		printf(temp==1?"Yes\n":"No\n");
    	}
    	return 0;
    }

[Status]: http://acm.hust.edu.cn/vjudge/contest/127946#status//E/0
[Practice]: http://acm.hust.edu.cn/vjudge/problem/115821
[CodeForces 510B]: http://acm.hust.edu.cn/vjudge/problem/115821/origin
[Center]: /images/20220718/d7a5615071e8425b835147e915ff3bf2.png