Codeforces Round #124 (Div. 1) C. Paint Tree（极角排序）

C. Paint Tree

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given a tree with n vertexes and n points on a plane, no three points lie on one straight line.

Your task is to paint the given tree on a plane, using the given points as vertexes.

That is, you should correspond each vertex of the tree to exactly one point and each point should correspond to a vertex. If two vertexes of the tree are connected by an edge, then the corresponding points should have a segment painted between them. The segments that correspond to non-adjacent edges, should not have common points. The segments that correspond to adjacent edges should have exactly one common point.

Input

The first line contains an integer n (1?≤?n?≤?1500) — the number of vertexes on a tree (as well as the number of chosen points on the plane).

Each of the next n?-?1 lines contains two space-separated integers u_i and v_i (1?≤?u_i,?v_i?≤?n, u_i?≠?v_i) — the numbers of tree vertexes connected by the i-th edge.

Each of the next n lines contain two space-separated integers x_i and y_i (?-?10⁹?≤?x_i,?y_i?≤?10⁹) — the coordinates of the i-th point on the plane. No three points lie on one straight line.

It is guaranteed that under given constraints problem has a solution.

Output

Print n distinct space-separated integers from 1 to n: the i-th number must equal the number of the vertex to place at the i-th point (the points are numbered in the order, in which they are listed in the input).

If there are several solutions, print any of them.

Examples

Input

31 32 30 01 12 0

Output

1 3 2

Input

41 22 31 4-1 -23 5-3 32 0

Output

4 2 1 3【分析】题意比较简单。给你一棵n个节点的树，再给你几何平面内的n个点，没有三点共线， 问你能不能用着n个点在几何平面内表现出来，线段除了顶点处无其他交点。 由于没有3点共线的情况，所以解总是存在的。 我们考虑以当前平面左下角的点p作为树根，对平面上的点以p做基准进行极角排序，则所有点与p点的连线都不会有除了p以外的交点。 现在我们已经会填树根处的点了，对于树根的每个子节点，我们都可以递归的处理以其为根的子树， 假设该子树包含x个节点，我们考虑以一根从p出发，长度不限的射线，从p的正下方开始按逆时针扫过去， 直到扫过的平面包含x个节点即可停止。此时扫过的平面即为该子树应当处理的平面。 每次处理需要先找到左下角的点，然后对当前平面的所有点进行排序，共需要处理n次，所以复杂度O(n^2*logn)。

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <string>
#include <stack>
#include <queue>
#include <vector>
#define inf 0x3f3f3f3f
#define met(a,b) memset(a,b,sizeof a)
#define pb push_back
typedef long long ll;
using namespace std;
const int N = 2005;
const int M = 24005;
pair<ll,ll>ori;
vector<ll>edg[N];
ll n,m,k;
ll sz[N],ans[N];
struct man{
    ll x,y,id;
    bool operator < (const man & b) const {
        return (x-ori.first)*(b.y-ori.second) - (y-ori.second)*(b.x-ori.first) > 0;
    }
}p[N];
void dfs1(ll u,ll fa){
    sz[u]=1;
    for(int i=0;i<edg[u].size();i++){
        ll v=edg[u][i];
        if(v==fa)continue;
        dfs1(v,u);
        sz[u]+=sz[v];
    }
}
void dfs2(ll u,ll fa,ll s,ll e){
    int pos;
    ll x,y;
    x=y=1e10;
    for(int i=s;i<=e;i++){
        if(p[i].x<x||((p[i].x==x)&&p[i].y<y)){
            x=p[i].x;y=p[i].y;pos=i;
        }
    }
    ori.first=x;ori.second=y;
    swap(p[s],p[pos]);
    sort(p+s+1,p+e+1);
    ans[p[s].id]=u;
    int cnt=0;
    for(int i=0;i<edg[u].size();i++){
        ll v=edg[u][i];
        if(v==fa)continue;
        dfs2(v,u,s+1+cnt,s+1+cnt+sz[v]-1);
        cnt+=sz[v];
    }
}
int main() {
    ll T,u,v;
    scanf("%lld",&n);
    for(int i=1;i<n;i++){
        scanf("%lld%lld",&u,&v);
        edg[u].pb(v);edg[v].pb(u);
    }
    for(int i=0;i<n;i++){
        scanf("%lld%lld",&p[i].x,&p[i].y);
        p[i].id=i;
    }
    dfs1(1,0);
    dfs2(1,0,0,n-1);
    for(int i=0;i<n;i++)printf("%lld ",ans[i]);printf("\n");
    return 0;
}