分治，选最左上的点分给根，剩下的极角排序后递归
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 ui and vi (1?≤?ui,?vi?≤?n, ui?≠?vi) ― the numbers of tree vertexes connected by the i-th edge.
Each of the next n lines contain two space-separated integers xi and yi (?-?109?≤?xi,?yi?≤?109) ― the coordinates of thei-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.
Sample test(s) input
3
1 3
2 3
0 0
1 1
2 0
output
1 3 2
input
4
1 2
2 3
1 4
-1 -2
3 5
-3 3
2 0
output
4 2 1 3
Note
The possible solutions for the sample are given below.
<??http://www.2cto.com/kf/ware/vc/" target="_blank" class="keylink">vcD4KPHA+PGJyPgo8L3A+CjxwPjxwcmUgY2xhc3M9"brush:java;">#include
#include
#include
#include
#include
using namespace std; const int maxn=1520; int n,X,Y; struct PO { int x,y,d; bool operator<(const PO &o) const { if(x-X>=0&&o.x-X<=0) return 1; if(x-X<=0&&o.x-X>=0) return 0; return (y-Y)*(long long)(o.x-X)<(o.y-Y)*(long long)(x-X); } }p[maxn]; vector
g[maxn]; bool vis[maxn]; int sz[maxn],o[maxn]; int dfs(int u) { vis[u]=true; sz[u]=1; int ret=0; for(int i=0,j=g[u].size();i