CF29D 樹的遍歷

 

D. Ant on the Tree time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output

Connected undirected graph without cycles is called a tree. Trees is a class of graphs which is interesting not only for people, but for ants too.

An ant stands at the root of some tree. He sees that there are n vertexes in the tree, and they are connected by n - 1 edges so that there is a path between any pair of vertexes. A leaf is a distinct from root vertex, which is connected with exactly one other vertex.

The ant wants to visit every vertex in the tree and return to the root, passing every edge twice. In addition, he wants to visit the leaves in a specific order. You are to find some possible route of the ant.

Input

The first line contains integer n (3 ≤ n ≤ 300) — amount of vertexes in the tree. Next n - 1 lines describe edges. Each edge is described with two integers — indexes of vertexes which it connects. Each edge can be passed in any direction. Vertexes are numbered starting from 1. The root of the tree has number 1. The last line contains k integers, where k is amount of leaves in the tree. These numbers describe the order in which the leaves should be visited. It is guaranteed that each leaf appears in this order exactly once.

Output

If the required route doesn't exist, output -1. Otherwise, output 2n - 1 numbers, describing the route. Every time the ant comes to a vertex, output it's index.

Sample test(s) input

3
1 2
2 3
3

output

1 2 3 2 1 

input

6
1 2
1 3
2 4
4 5
4 6
5 6 3

output

1 2 4 5 4 6 4 2 1 3 1 

input

6
1 2
1 3
2 4
4 5
4 6
5 3 6

output

-1

/**
CF29D 樹的遍歷
題目大意：
         給定一個樹以1為根節點，給定每個葉子節點的遍歷順序，從1出發按照此順序遍歷最後再回到1，
         經過沒條邊2次，問是否可以，若可以輸出之
解題思路：
        這是一棵樹（無向無環連通圖），那麼我們遍歷的時候假設一直的葉子節點分別為a1,a2,a3，
        那麼我們先dfs出1~a1之間的路，接著a1~a2，a2~a3,a3~1,過程中統一放在一個容器中，最後若正好
        是2*n-1個點加入容器，那麼就有解（無向無環圖，點u到點v的路徑如果有那就只有一條，說明這個
        方法可行），輸出就好了
*/
#include 
#include 
#include 
#include 
#include 
using namespace std;

int n;
vector vec[500],a;

bool dfs(int root,int u,int pre)
{
    if(root==u)return 1;
    for(int i=0; i