codeforces 29D Ant on the Tree (dfs,tree,最近公共祖先）

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

 题意：给你一个无向图，先变成以1为根的树，在判断是否能用给的顺序访问叶子节点，每条边只走两次；

思路：先dfs变成tree，在找给的顺序相邻两叶子的最近公共祖先，把路径存下来，最后判断存下的路径里的节点是不是2*n-1个，是就输出路径；

AC代码：

#include <bits/stdc++.h>
using namespace std;
vector<int>v[302];
vector<int>ans;
stack<int>sta;
queue<int>qu;
int c[300],flag[303],fa[303];
int lca(int x,int y)
{
    int start=x;
    memset(flag,0,sizeof(flag));
    flag[x]=1;
    while(x!=1)
    {
        x=fa[x];
        flag[x]=1;
    }
    while(!flag[y])
    {
        sta.push(y);
        y=fa[y];
    }
    while(start!=y)
    {
        start=fa[start];
        ans.push_back(start);
    }
    while(!sta.empty())
    {
        ans.push_back(sta.top());
        sta.pop();
    }
}
int bfs()
{
    memset(flag,0,sizeof(flag));
    while(!qu.empty()){
    int fr=qu.front();
    qu.pop();
    int len=v[fr].size();
    for(int i=0;i<len;i++)
    {
        if(flag[v[fr][i]]==0)
        {
        fa[v[fr][i]]=fr;
        qu.push(v[fr][i]);
        flag[v[fr][i]]=1;
        }
    }
    }
}
int main()
{
    int n,a,b;
    scanf("%d",&n);
    for(int i=0;i<n-1;i++)
    {
        scanf("%d%d",&a,&b);
        v[a].push_back(b);
        v[b].push_back(a);
    }
    qu.push(1);
    bfs();
    int k=0;
    for(int i=2;i<=n;i++)
    {
        if(v[i].size()==1)k++;
    }
    for(int i=1;i<=k;i++)
    {
        scanf("%d",&c[i]);
    }
    ans.push_back(1);
    c[k+1]=1;
    c[0]=1;
    for(int i=0;i<=k;i++)
    {
        lca(c[i],c[i+1]);
    }
    int len=ans.size();
    if(len!=2*n-1)
    {
        cout<<"-1"<<endl;
        return0;
　　}
　　for(int i=0;i<len;i++)printf("%d ",ans[i]);
return0;
}

求最近公共祖先用的最原始的方法，其实可以优化；还有就是最近一直喜欢用这些容器啥的，还是太笨啊！