John’s trip
| **Time Limit:** 1000MS | **Memory Limit:** 65536K | |||
| **Total Submissions:** 3223 | **Accepted:** 952 | 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 =
- 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<br></br>2 3 2<br></br>3 1 6<br></br>1 2 5<br></br>2 3 3<br></br>3 1 4<br></br>0 0<br></br>1 2 1<br></br>2 3 2<br></br>1 3 3<br></br>2 4 4<br></br>0 0<br></br>0 0
Sample Output
1 2 3 5 4 6 <br></br>Round trip does not exist.<br></br><span style="font-family: Arial;"><br></br>最近几天基本都没做题目……<br></br>这题目那天看了感觉是欧拉回路，判断还好说，输出一条字典序最小的路感觉有点麻烦。搜索了一下输出欧拉回路的方法，发现基本思想是先找一条回路，然后在找到的回路中以度不为0的点为起点继续找回路，插入到一开始找到的回路中。直到所有边都被枚举。<br></br>看了这个思路感觉想不出什么方便的办法……开两个过程一个递归，然后另外一个再处理插入啥的……为了减小复杂度还用到链表……各种麻烦……<br></br>百度里面的算法很简单，一开始一直没看懂，后来突然一下懂了，先枚举的点放在后面，之后倒序输出，相当棒的算法……而且貌似曾经我知道这玩意儿……而且根据这个算法只要循环的时候枚举边就自动输出字典序……题目还是比较简单的居然……<br></br>贴代码……<br></br>#include <stdio.h><br></br>#include <string.h><br></br>#include <list><br></br><br></br>using namespace std;<br></br><br></br>typedef list<int> listInt;<br></br>listInt ans;<br></br><br></br>int n = 2000, m = 45;<br></br>int t;<br></br>int street[2010][3];<br></br>int match[2010];<br></br><br></br>int start;<br></br><br></br>bool Init()<br></br>{<br></br>int x, y, z, tmp;<br></br>memset(street, 0, sizeof(street));<br></br>memset(match, 0, sizeof(match));<br></br>ans.clear();<br></br>scanf("%d%d", &x, &y);<br></br>if(x == 0 && y == 0)<br></br>return false;<br></br>start = x;<br></br>if(start > y)<br></br>start = y;<br></br>while(x != 0){<br></br>scanf("%d", &z);<br></br>if(x > y){<br></br>tmp = x;<br></br>x = y;<br></br>y = tmp;<br></br>}<br></br>match[x]++;<br></br>match[y]++;<br></br>street[z][0] = x;<br></br>street[z][1] = y;<br></br>scanf("%d%d", &x, &y);<br></br>}<br></br>return true;<br></br>}<br></br><br></br>int father[50];<br></br><br></br>int GetFar(int k)<br></br>{<br></br>if(father[k] == k)<br></br>return k;<br></br>else{<br></br>father[k] = GetFar(father[k]);<br></br>return father[k];<br></br>}<br></br>}<br></br><br></br>bool Check()<br></br>{<br></br>for(int i=0; i<=m; i++)<br></br>father[i] = i;<br></br>for(int i=1; i<=n; i++){<br></br>if(street[i][0] != 0 && street[i][1] != 0){<br></br>father[GetFar(street[i][0])] = GetFar(street[i][1]);<br></br>}<br></br>}<br></br>for(int i=1; i<=n; i++){<br></br>if(match[i] == 0)<br></br>continue;<br></br>if(match[i] % 2 == 1)<br></br>return false;<br></br>if(GetFar(i) != GetFar(start))<br></br>return false;<br></br>}<br></br>return true;<br></br>}<br></br><br></br>void Euler(int start)<br></br>{<br></br>if(match[start] > 0){<br></br>for(int i=1; i<=n; i++){<br></br>if(street[i][2] == 0 &&(street[i][0] == start || street[i][1] == start)){<br></br>street[i][2] = 1;<br></br>match[street[i][0]] --;<br></br>match[street[i][1]] --;<br></br>if(street[i][0] == start){<br></br>Euler(street[i][1]);<br></br>}else if(street[i][1] == start){<br></br>Euler(street[i][0]);<br></br>}<br></br>ans.push_back(i);<br></br>}<br></br>}<br></br>}<br></br>}<br></br><br></br>void Solve()<br></br>{<br></br>if(!Check()){<br></br>printf("Round trip does not exist.\n");<br></br>return ;<br></br>}<br></br>t = 0;<br></br>Euler(start);<br></br><br></br>for(listInt::const_reverse_iterator i = ans.rbegin(); i != ans.rend(); ++i){<br></br>if(i != ans.rbegin()){<br></br>printf(" ");<br></br>}<br></br>printf("%d", *i);<br></br>}<br></br>printf("\n");<br></br>}<br></br><br></br>int main()<br></br>{<br></br>while(Init()){<br></br>Solve();<br></br>}<br></br>}<br></br><br></br><br></br></span>