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:
- These k dots are different: if i ≠ j then di is different from dj.
- k is at least 4.
- All dots belong to the same color.
- 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
3 4
AAAA
ABCA
AAAA
Yes
3 4
AAAA
ABCA
AADA
No
4 4
YYYR
BYBY
BBBY
BBBY
Yes
7 6
AAAAAB
ABBBAB
ABAAAB
ABABBB
ABAAAB
ABBBAB
AAAAAB
Yes
2 13
ABCDEFGHIJKLM
NOPQRSTUVWXYZ
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).
题目大意 :判断图中是否有些相同字母组成的环,如果有的话直接出书YES,没有输出NO
思路 : 图中的每个点都有可能构成循环,所以要逐一遍历,如果该点可以构成循环的环 那么起点是该点,,终点也是该点,环至少是4个点构成,所以如果有环的话再次回到起点时步数一定大于等于四。
AC代码:
#include<iostream> #include<cstdio> #include<cstring> using namespace std; int n,m,end_i,end_j; char aa; char arr[100][100]; int mark[100][100]={0}; int d[4][2]={{1,0},{-1,0},{0,1},{0,-1}}; int flag=0; void dfs(int x,int y,int step) { if(flag==1) return; if(step>=4&&x==end_i&&y==end_j)如果回到了起点且步数大于等于4 则找到环了。 { flag=1; return; } for(int i=0;i<4;i++){ int dx=x+d[i][0]; int dy=y+d[i][1]; if(dx>=0&&dy>=0&&dx<n&&dy<m&&mark[dx][dy]==0&&arr[dx][dy]==aa){ if(dx==end_i&&y==end_j&&step<3)//起点就是终点所以起点没有被标记,当进入到第二个点时,可能会回到起点,所以要对步数进行判断,看是否大于等于3 continue; mark[dx][dy]=1; dfs(dx,dy,step+1); } } } int main() { cin>>n>>m;//n行m列 for(int i=0;i<n;i++) scanf("%s",&arr[i]);//存图 for(int i=0;i<n;i++) { for(int j=0;j<m;j++) { memset(mark,0,sizeof(mark)); end_i=i; end_j=j;//起点与终点 aa=arr[i][j];//搜索的字符 dfs(i,j,0); if(flag==1) break; } if(flag==1) break; } if(flag==1) cout<<"Yes"<<endl; else cout<<"No"<<endl; return 0; }