1 问题描述
John’s trip
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 8998 Accepted: 3018 Special Judge
Description
Little Johnny has got a new car. He decided to drive around the town to visit his friends. Johnny wanted to visit all his friends, but there was many of them. In each street he had one friend. He started thinking how to make his trip as short as possible. Very soon he realized that the best way to do it was to travel through each street of town only once. Naturally, he wanted to finish his trip at the same place he started, at his parents’ house.
The streets in Johnny’s town were named by integer numbers from 1 to n, n < 1995. The junctions were independently named by integer numbers from 1 to m, m <= 44. No junction connects more than 44 streets. All junctions in the town had different numbers. Each street was connecting exactly two junctions. No two streets in the town had the same number. He immediately started to plan his round trip. If there was more than one such round trip, he would have chosen the one which, when written down as a sequence of street numbers is lexicographically the smallest. But Johnny was not able to find even one such round trip.
Help Johnny and write a program which finds the desired shortest round trip. If the round trip does not exist the program should write a message. Assume that Johnny lives at the junction ending the street appears first in the input with smaller number. All streets in the town are two way. There exists a way from each street to another street in the town. The streets in the town are very narrow and there is no possibility to turn back the car once he is in the street
Input
Input file consists of several blocks. Each block describes one town. Each line in the block contains three integers x; y; z, where x > 0 and y > 0 are the numbers of junctions which are connected by the street number z. The end of the block is marked by the line containing x = y = 0. At the end of the input file there is an empty block, x = y = 0.
Output
Output one line of each block contains the sequence of street numbers (single members of the sequence are separated by space) describing Johnny’s round trip. If the round trip cannot be found the corresponding output block contains the message “Round trip does not exist.”
Sample Input
1 2 1
2 3 2
3 1 6
1 2 5
2 3 3
3 1 4
0 0
1 2 1
2 3 2
1 3 3
2 4 4
0 0
0 0
Sample Output
1 2 3 5 4 6
Round trip does not exist.
Source
Central Europe 1995
中文翻译
约翰的旅行
时间限制：1000毫秒内存限制：65536K
提交总数：8998件接受人数：3018名特别法官
说明
小约翰尼有辆新车。他决定开车环城去拜访他的朋友。约翰尼想拜访他所有的朋友，但他们中有很多人。他在每条街上都有一个朋友。他开始考虑如何尽可能缩短行程。很快他就意识到，最好的办法是只穿过城镇的每一条街道一次。当然，他想在他开始的同一个地方完成他的旅行，在他父母的家里。
约翰尼镇上的街道是用1到n的整数命名的，n<1995。连接点由1到m的整数独立命名，m<=44。没有连接44条以上街道的交叉口。镇上所有的路口都有不同的号码。每条街道正好连接两个路口。镇上没有两条街道的号码相同。他立即开始计划他的往返旅行。如果有不止一次这样的往返旅行，他会选择一次，当写下街道号码的序列时，从字典上来说，是最小的。但约翰尼甚至找不到这样的往返旅行。
帮助约翰尼写一个程序，找出最短的往返行程。如果往返不存在，程序应编写一条消息。假设约翰尼住在街尾的交叉路口，在输入中以较小的数字出现在第一位。镇上所有的街道都是双向的。镇上有一条从每条街道到另一条街道的路。城里的街道很窄，一旦他在街上，就不可能把车掉头。
输入
输入文件由几个块组成。每个街区都有一个城镇。块中的每一行包含三个整数x；y；z，其中x>0和y>0是由街道编号z连接的连接数。块的结尾由包含x=y=0的行标记。在输入文件的末尾有一个空块，x=y=0。
输出
输出每个街区的一行包含描述约翰尼往返路线的街道编号序列（序列中的单个成员用空格分隔）。如果找不到往返行程，则相应的输出块包含消息“往返不存在”。
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 = 100; //题意说明，最多44个路口
public static int start; // Johnny出发起点
public static int[] id = new int[MAX];
public static int[] degree = new int[MAX]; //用于计算给定图每个顶点的度
public static boolean[] used = new boolean[2000]; //用于判断图中相应边是否被遍历
public static int[] path = new int[MAX];
public static int count; //用于统计行走路径的街道数
public static ArrayList<String> result = new ArrayList<String>();
class MyComparator implements Comparator<edge> {
public int compare(edge o1, edge o2) {
if(o1.street > o2.street)
return 1;
else if(o1.street < o2.street)
return -1;
else
return 0;
}
}
static class edge {
public int a; //边的起点
public int b; //边的终点
public int street; //街道
public edge(int a, int b, int street) {
this.a = a;
this.b = b;
this.street = street;
}
}
//寻找顶点a的根节点
public int find(int[] id, int a) {
int root = a;
while(id[root] >= 0) {
root = id[root];
}
int i;
int k = a;
while(k != root) {
i = id[k];
id[k] = root;
k = i;
}
return root;
}
//合并顶点a和顶点b所在的树
public void union(int[] id, int a, int b) {
int rootA = find(id, a);
int rootB = find(id, b);
int rootNum = id[rootA] + id[rootB];
if(id[rootA] < id[rootB]) {
id[rootB] = rootA;
id[rootA] = rootNum;
} else{
id[rootA] = rootB;
id[rootB] = rootNum;
}
return;
}
public void init() {
for(int i = 0;i < 50;i++) {
id[i] = -1; //初始化所有顶点所在树的根节点编号为-1
degree[i] = 0;
path[i] = -1;
count = 0;
}
for(int i = 0;i < 2000;i++) {
used[i] = false;
}
return;
}
public boolean judge(ArrayList<edge>[] map) {
int root = find(id, start);
for(int i = 0;i < MAX;i++) {
if(map[i].size() == 0)
continue;
Collections.sort(map[i], new MyComparator());
for(int j = 0;j < map[i].size();j++) {
if(find(id, map[i].get(j).b) != root)
return false;
}
}
for(int i = 0;i < MAX;i++) {
if(degree[i] % 2 != 0)
return false;
}
return true;
}
public void dfs(ArrayList<edge>[] map, int start) {
for(int i = 0;i < map[start].size();i++) {
if(!used[map[start].get(i).street]) {
used[map[start].get(i).street] = true;
path[count++] = map[start].get(i).street;
dfs(map, map[start].get(i).b);
}
}
}
public static void main(String[] args) {
Main test = new Main();
Scanner in = new Scanner(System.in);
while(true) {
int a1 = in.nextInt();
int b1 = in.nextInt();
if(a1 == 0 && b1 == 0)
break;
int c1 = in.nextInt();
start = Math.min(a1, b1); //Johnny出发起点
test.init(); //初始化输入顶点的度和所在树的根节点
@SuppressWarnings("unchecked")
ArrayList<edge>[] map = new ArrayList[MAX];
for(int i = 0;i < MAX;i++) {
map[i] = new ArrayList<edge>();
}
map[a1].add(new edge(a1, b1, c1));
map[b1].add(new edge(b1, a1, c1));
degree[a1]++;
degree[b1]++;
test.union(id, a1, b1); //合并顶点a1和顶点b1所在树
while(true) {
int a = in.nextInt();
int b = in.nextInt();
if(a == 0 && b == 0)
break;
int c = in.nextInt();
map[a].add(new edge(a, b, c));
map[b].add(new edge(b, a, c));
degree[a]++;
degree[b]++;
test.union(id, a, b);
}
String tempR = "";
if(test.judge(map)) {
test.dfs(map, start);
for(int i = 0;i < count;i++) {
tempR = tempR + path[i] + " ";
}
} else {
tempR = "Round trip does not exist.";
}
result.add(tempR);
}
for(int i = 0;i < result.size();i++)
System.out.println(result.get(i));
}
}
运行结果：
2 1
3 2
1 6
2 5
3 3
1 4
0
2 1
3 2
3 3
4 4
0
0
2 3 5 4 6
Round trip does not exist.
