分治。选最左上的点分给根。剩下的极角排序后递归
2 seconds
256 megabytes
standard input
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.
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.
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.
3
1 3
2 3
0 0
1 1
2 0
1 3 2
4
1 2
2 3
1 4
-1 -2
3 5
-3 3
2 0
4 2 1 3
The possible solutions for the sample are given below.
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector> 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<int> 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<j;i++)
{
int v=g[u][i];
if(vis[v]) continue;
ret+=dfs(v);
}
sz[u]+=ret;
return sz[u];
} void calc(int u,int l,int r)
{
vis[u]=true;
int t=l;
for(int i=l+1;i<=r;i++)
{
if((p[i].y<p[t].y)||(p[t].y==p[i].y&&p[i].x<p[t].x))
t=i;
}
if(t!=l) swap(p[l],p[t]);
o[p[l].d]=u;
X=p[l].x; Y=p[l].y;
sort(p+l+1,p+r+1);
int pos=l+1;
for(int i=0,j=g[u].size();i<j;i++)
{
int v=g[u][i];
if(vis[v]) continue;
calc(v,pos,pos+sz[v]-1);
pos+=sz[v];
}
}
int main()
{
scanf("%d",&n);
for(int i=0;i<n-1;i++)
{
int x,y;
scanf("%d%d",&x,&y);
g[x].push_back(y); g[y].push_back(x);
}
for(int i=1;i<=n;i++)
{
int x,y;
scanf("%d%d",&x,&y);
p[i]=(PO){x,y,i};
}
dfs(1);
memset(vis,0,sizeof(vis));
calc(1,1,n);
for(int i=1;i<=n;i++)
printf("%d ",o[i]);
putchar(10);
return 0;
}