codeforces-Fox And Two Dots【DFS】（思维）

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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:

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 - 1: di 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.

Examples

input

3 4AAAAABCAAAAA

output

Yes

input

3 4AAAAABCAAADA

output

No

input

4 4YYYRBYBYBBBYBBBY

output

Yes

input

7 6AAAAABABBBABABAAABABABBBABAAABABBBABAAAAAB

output

Yes

input

2 13ABCDEFGHIJKLMNOPQRSTUVWXYZ

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).

题意：在n*m的地图上，每个格子涂有不同的颜色（A~Z），问是否有某一种颜色能组成一个回路且长度至少为4. 

#include<cstdio>#include<algorithm>#include<cstring>using namespace std;int n,m;char map[110][110];bool vis[110][110];int dx[4]={-1,1,0,0};int dy[4]={0,0,-1,1};bool judge(int x,int y){if(x>=0&&x<n&&y>=0&&y<m)return 1;return 0;}bool DFS(int x,int y,int px,int py,int cnt){vis[x][y]=1;for(int i=0;i<4;i++){int nx1=x+dx[i];int ny1=y+dy[i];if(vis[nx1][ny1]&&judge(nx1,ny1)&&(nx1!=px||ny1!=py)&&cnt>3) // 里面第三个判断条件是保证不会回头计算上一步刚走过的那个点return 1;}char c=map[x][y]; for(int i=0;i<4;i++){int nx2=x+dx[i];int ny2=y+dy[i];if(!vis[nx2][ny2]&&judge(nx2,ny2)&&map[nx2][ny2]==c){return DFS(nx2,ny2,x,y,cnt+1); // 这个不是 void 型的 DFS 所以调用自己要加 return }}return 0;}int main(){while(~scanf("%d %d",&n,&m)){bool flag=0;for(int i=0;i<n;i++)scanf("%s",map[i]);for(int i=0;i<n;i++){for(int j=0;j<m;j++){memset(vis,0,sizeof(vis));if( DFS(i,j,i,j,1) ){flag=1;break;}}if(flag)break;}if(flag)printf("Yes\n");elseprintf("No\n");}return 0;}