Codeforces-----510B Fox And Two Dots 搜索

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 sizen × 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 dotsd₁, d₂, ..., d_k acycle if and only if it meets the following condition:

1. These k dots are different: if i ≠ j then d_i is different fromd_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 andd₁ should also be adjacent. Cellsx 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 andm (2 ≤ n, m ≤ 50): the number of rows and columns of the board.

Then n lines follow, each line contains a string consisting ofm 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

给一幅图，问是否能构成环，没想到什么好的方法，暴力过，每个点都搜索一遍看能不能从这点出发再回到这个点路程不小于4

#include
#include
#include
#include
#define maxn 100010
#define inf 0x3f3f3f3f
#define ll long long
using namespace std;
int map[55][55];
bool vis[55][55];
int mov[4][2] = {0, 1, 0, -1, 1, 0, -1, 0};
int ans, ok, n, m, s, e;
void dfs(int x, int y){
    if(ok){
        return ;
    }
    if(vis[x][y]){
        if(ans >= 4 && x == s && y == e){//只需检查是否从一个点出发再次回到这个点并且路程不小于4
            ok++;
        }
        return ;
    }
    vis[x][y] = true;
    for(int i = 0; i < 4; i++){
        if(map[x+mov[i][0]][y+mov[i][1]] != map[x][y]){
            continue;
        }
        ans++;
        dfs(x+mov[i][0], y+mov[i][1]);
        ans--;
    }
    return ;
}
int main(){
	int t, kcase = 1;
	char ch;
	while(~scanf("%d%d", &n, &m)){
        memset(map, -1, sizeof(map));
        for(int i = 1; i <= n; i++){
            for(int j = 1; j <= m; j++){
                scanf(" %c", &ch);
                map[i][j] = ch-64;//便于判断边界，直接memset初始化-1
            }
        }
        ok = 0;
        for(int i = 1; i <= n; i++){
            for(int j = 1; j <= m; j++){
                memset(vis, false, sizeof(vis));
                ans = 1;
                s = i;
                e = j;
                dfs(i, j);
                if(ok){
                    break;
                }
            }
            if(ok){
                break;
            }
        }
        printf(ok ? "Yes\n" : "No\n");
	}
	return 0;
}