题意

给出一系列字符串，找出一列最长的字符串队列满足：

• A比B少一个字母c
• 除了字母c外，A中所有字母都包含于字母B中。

思路

前面思路想直接暴力模拟，后来发现行不通，需要用到DFS，想到DFS后就是构树，这些就比较简单了。还有C++中还有Hash和二分法之类的。

注意

DFS递归时要搞清Java的值传递和引用传递！！！

代码

import java.io.PrintWriter;
import java.util.*;

/**
 * @Author Yuri
 * @Date 2020/11/26 10:22
 * @Version 1.0
 * @Description: POJ2004
 */

class Letters implements Comparable<Letters> {
    String sortName, name;
    ArrayList<Integer> children = new ArrayList<Integer>();
    boolean flag;
    Letters(String sortName, String name) {
        this.sortName = sortName;
        this.name = name;
        flag = true;
    }


    @Override
    public int compareTo(Letters o) {
        return name.length() - o.name.length();
    }

    @Override
    public String toString() {
        return "Letters{" +
                "sortName='" + sortName + '\'' +
                ", name='" + name + '\'' +
                ", children=" + children +
                ", flag=" + flag +
                '}';
    }
}

public class Main {
    static PrintWriter out = new PrintWriter(System.out);
    static Scanner scanner = new Scanner(System.in);
    static Integer n = 0;
    static ArrayList<Letters> letters = new ArrayList<Letters>();


    static ArrayList<Integer> ans = new ArrayList<Integer>();


    public static void stringSort(String s) {
        char[] c = s.toCharArray();
        Arrays.sort(c);
        String ss = new String(c);
        letters.add(new Letters(ss, s));
    }

    public static boolean isOK(String x, String y) {
        int i = 0;
        while (i < x.length() && x.charAt(i) == y.charAt(i)) i++;
        while (++i < y.length()) {
            if (y.charAt(i) != x.charAt(i - 1))
                return false;
        }
        return true;
    }

    public static void dfs(Integer r, ArrayList<Integer> fakeAns) {
        Letters lt = letters.get(r);
        ArrayList<Integer> fa = new ArrayList<Integer>(fakeAns);
        fa.add(r);
        if (lt.children.isEmpty()){
            if (ans.size() < fa.size()) {
                ans = fa;
            }
            return;
        }
        for (int i = 0; i < lt.children.size(); i++) {
            dfs(lt.children.get(i), fa);
        }
    }

    public static void main(String[] args) {
        while (scanner.hasNext()) {
            n++;
            String s = scanner.next();
            stringSort(s);
        }
        Collections.sort(letters);
        for (int i = 0; i < n; i++) {
            for (int j = i + 1; j < n; j++) {
                String si = letters.get(i).sortName;
                String sj = letters.get(j).sortName;
                if (si.length() + 1 == sj.length() && isOK(si, sj)) {
                    letters.get(i).children.add(j);
                    letters.get(j).flag = false;
                } else if (si.length() + 1 < sj.length()) {
                    break;
                }
            }
        }
        for (int i = 0; i < n; i++) {
            ArrayList<Integer> fakeAns = new ArrayList<Integer>();
            if (letters.get(i).flag) {
                dfs(i, fakeAns);
            }
        }
        for (int i = 0; i < ans.size(); i++) {
            System.out.println(letters.get(ans.get(i)).name);
        }
    }
}