标签：queue printf tle ext segment 开始 not tac com
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 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.
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
【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; }
Codeforces Round #124 (Div. 1) C. Paint Tree（极角排序）
标签：queue printf tle ext segment 开始 not tac com
原文地址：http://www.cnblogs.com/jianrenfang/p/6528122.html
