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
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.
题目大意：john要去拜访他的朋友，要经过每一条街道，问是否存在这样的路径
题解 ：要经过每一条边，就是欧拉回路，欧拉回路存在的充要条件是：无向图中节点的度为偶数，有向图中节点的入度等于出度
代码是仿照别人写的，以前没碰到过这种题
/*poj1041 第一次看到还以为是flord最小环问题，结果发现是欧拉回路上网看了一下大佬们的思路 然后自己在写一遍代码，加深一下印象 欧拉回路：每一条边都要经过，充要条件是：无向图的节点度为偶数个 有向图的节点入度等于出度*/ #include<iostream> #include<cstring> #include<algorithm> #include<cstdio> using namespace std; int street_cnt;//街道的个数 int street[1996][2];//街道链接的两端 int map[45][1995];//map[i][j]表示从i点经过j点到达的路径 bool visited[1996];//判断有没有访问过 int degree[45];//存储节点的入度 int Stack[1995];//用一个栈来存放路径 int stack_top;//来表示栈头元素 int home; void record(int x,int y,int z) { street[z][0]=x;//街道两端链接的家 street[z][1]=y; map[x][z]=y;//通过这条街道可以到达的朋友家 map[y][z]=x; ++degree[x];//然后节点度数要增加 ++degree[y]; } void DFS(int j) { for(int i=1;i<=street_cnt;i++) { if(!visited[i]&&map[j][i]) { visited[i]=true; DFS(map[j][i]);//到了当前的朋友家然后开始下一个朋友 Stack[stack_top++]=i; } } } int main() { while (true) { int x, y, z; scanf("%d%d", &x, &y); if (x == 0 && y == 0) { break; } scanf("%d", &z); memset(map, false, sizeof(map)); memset(degree, 0, sizeof(degree)); record(x, y, z); home = x < y ? x : y; street_cnt = 1; while (true) { scanf("%d%d", &x, &y); if (x == 0 && y == 0) { break; } scanf("%d", &z); record(x, y, z); ++street_cnt; } bool flag = true; //欧拉回路存在的充要条件是每个顶点的度数都为偶数 for (int i = 1; i <= 44; ++i) { if (degree[i] % 2 != 0) { flag = false; break; } } if (flag == false) { printf("Round trip does not exist. "); continue; } memset(visited, false, sizeof(visited)); stack_top = 0; DFS(home); for (int i = stack_top - 1; i > 0; --i) { printf("%d ", Stack[i]); } printf("%d ", Stack[0]); } return 0; }