Codeforces Round #290 (Div. 2) B. Fox And Two Dots dfs-蒲公英云

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

梦里梦外; 2022-09-25 05:18 233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 d1, d2, …, d**k a cycle if and only if it meets the following condition:

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

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

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;
}