目录
1 问题描述
2 解决方案
1 问题描述
Description
A catenym is a pair of words separated by a period such that the last letter of the first word is the same as the last letter of the second. For example, the following are catenyms:
dog.gopher gopher.rat rat.tiger aloha.aloha arachnid.dog
A compound catenym is a sequence of three or more words separated by periods such that each adjacent pair of words forms a catenym. For example,
aloha.aloha.arachnid.dog.gopher.rat.tiger
Given a dictionary of lower case words, you are to find a compound catenym that contains each of the words exactly once.
Input
The first line of standard input contains t, the number of test cases. Each test case begins with 3 <= n <= 1000 - the number of words in the dictionary. n distinct dictionary words follow; each word is a string of between 1 and 20 lowercase letters on a line by itself.
Output
For each test case, output a line giving the lexicographically least compound catenym that contains each dictionary word exactly once. Output "***" if there is no solution.
Sample Input
2 6 aloha arachnid dog gopher rat tiger 3 oak maple elm
Sample Output
aloha.arachnid.dog.gopher.rat.tiger ***
Source
题目链接：http://poj.org/problem?id=2337
2 解决方案
具体代码如下：
package com.liuzhen.practice;
import java.util.ArrayList;
import java.util.Collections;
import java.util.Comparator;
import java.util.Scanner;
public class Main {
public static int MAX = 30; //英文字母共26个
@SuppressWarnings("unchecked")
public static ArrayList<edge>[] map = new ArrayList[MAX];
public static String[] path;
public static int count; //统计DFS遍历访问的边数目，即检测图的连通性
public static int start;
public static int[] inDegree = new int[MAX];
public static int[] outDegree = new int[MAX];
public static ArrayList<String> result = new ArrayList<String>();
class MyComparator implements Comparator<edge> {
public int compare(edge arg0, edge arg1) {
String A = arg0.word;
String B = arg1.word;
int judge = A.compareTo(B);
if(judge > 0)
return 1;
else if(judge < 0)
return -1;
return 0;
}
}
static class edge {
public int a; //单词的第一个字母序号
public int b; //单词最后一个字母序号
public String word; //具体单词
public boolean used; //判断单词是否被访问
public edge(int a, int b, String word) {
this.a = a;
this.b = b;
this.word = word;
used = false;
}
}
public void init(int k) {
start = MAX;
for(int i = 0;i < MAX;i++) {
map[i] = new ArrayList<edge>();
inDegree[i] = 0;
outDegree[i] = 0;
}
path = new String[k];
for(int i = 0;i < k;i++)
path[i] = "";
count = 0;
}
public boolean judgeDegree() {
int in = 0, out = 0;
for(int i = 1;i < map.length;i++) { //对map[i]中单词进行字典序排序
if(map[i].size() > 1)
Collections.sort(map[i], new MyComparator());
}
for(int i = 0;i < inDegree.length;i++) {
if(inDegree[i] == outDegree[i])
continue;
else if(inDegree[i] - outDegree[i] == 1)
in++;
else if(outDegree[i] - inDegree[i] == 1) {
out++;
start = i; //此时，可能存在欧拉路径，必须从入度小于出度的点开始遍历
} else
return false;
}
if(in == out && (in == 0 || in == 1))
return true;
return false;
}
public void dfs(int begin) {
for(int i = 0;i < map[begin].size();i++) {
edge temp = map[begin].get(i);
if(temp.used == false) {
temp.used = true;
path[count++] = temp.word;
dfs(temp.b);
}
}
}
public static void main(String[] args) {
Main test = new Main();
Scanner in = new Scanner(System.in);
int t = in.nextInt();
while(t > 0) {
t--;
int k = in.nextInt();
test.init(k);
for(int i = 0;i < k;i++) {
String A = in.next();
int a = A.charAt(0) - ‘a‘;
int b = A.charAt(A.length() - 1) - ‘a‘;
start = Math.min(start, Math.min(a, b));
map[a].add(new edge(a, b, A));
outDegree[a]++;
inDegree[b]++;
}
StringBuilder temp = new StringBuilder("");
if(test.judgeDegree()) { //满足欧拉回路或者欧拉路径对顶点度的要求
test.dfs(start);
if(count == k) { //图连通
for(int i = 0;i < k;i++) {
temp.append(path[i]);
if(i != k - 1)
temp.append(".");
}
} else {
temp.append("***");
}
} else {
temp.append("***");
}
result.add(temp.toString());
}
for(int i = 0;i < result.size();i++)
System.out.println(result.get(i));
}
}
运行结果：
2 6 aloha arachnid dog gopher rat tiger 3 oak maple elm aloha.arachnid.dog.gopher.rat.tiger ***
参考资料：
1. 欧拉回路