【DFS】困难的串

 1 import java.util.Scanner;
 2 
 3 public class 困难的串 {
 4 
 5     public static void main(String[] args) {
 6 //        int n = 10;
 7 //        int L = 4;
 8         Scanner scanner = new Scanner(System.in);
 9         int L = scanner.nextInt();
10         int n = scanner.nextInt();
11         dfs(L, n, "");
12         // isHard("0123020120",1);
13     }
14 
15     static int count;
16 
17     private static void dfs(int l, int n, String prefix) {
18 
19         // 尝试在prefix后追加一个字符  难点一：维持字典序
20         for (char i = ‘A‘; i < ‘A‘ + l; i++) {
21             if (isHard(prefix, i)) {// 是困难的串,就组合起来输出
22                 String x = prefix + i;
23                 //System.out.println(x);
24                 count++;// 计数
25                 if (count == n){
26                     System.out.println(x);
27                     System.exit(0);
28                 }
29 
30                 dfs(l, n, x);
31             }
32         }
33     }
34 
35     /**
36      *  难点二：判断prefix+i是否一个困难的串
37      *  1.遍历所有的长度为偶数的子串,看是否对称 
38      *  2.prefix本身是一个困难的串 ABACA i 往后遍历逐步比较
39      * @param prefix
40      * @param i
41      * @return
42      */
43     private static boolean isHard(String prefix, char i) {
44         int count = 0;// 截取的宽度
45         for (int j = prefix.length() - 1; j >= 0; j -= 2) {
46             final String s1 = prefix.substring(j, j + count + 1);
47             final String s2 = prefix.substring(j + count + 1) + i;
48             if (s1.equals(s2))
49                 return false;
50             count++;
51         }
52         return true;
53     }
54 }