[CF196E](Paint Tree)

solution from here

题意:

给你一颗n个点的无根树,及平面上的n个点,将它们一一对应,不允许线段交叉.

随便输出一个解,题目保证没有三点共线.

 

 

solution:
思路:极角排序+dfs分治

函数solve(l,r,x,fa)

l r为x及其子树对应的平面坐标区间

对于一棵树,把l~r中最左下角的那个点作为根

对剩下的点极角排序,

对于每一棵子树,按需分配－－即分配等同于其大小的点数

如上

 

code

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#define N 2000
#define ll long long
using namespace std;
ll n;
struct sett{
    ll x,y,id;
  }p[N],plk;
struct zero{
    ll nxt,to;
  }edge[N<<1];
ll tot=0,head[N];
ll size[N];
ll Ans[N];
void add_edge(ll a,ll b){
  edge[++tot]=(zero){head[a],b};
  head[a]=tot;
  }
void aux(ll x,ll fa){//预处理子树大小
    size[x]=1;
    for(ll i=head[x];i;i=edge[i].nxt){
        ll to=edge[i].to;
        if(to==fa)continue;
        aux(to,x);
        size[x]+=size[to];
        }
    }
bool cmp(sett a,sett b){
    ll ax=a.x-plk.x,ay=a.y-plk.y,bx=b.x-plk.x,by=b.y-plk.y;
    if(ax>=0&&bx<=0)return 1;
    if(ax<=0&&bx>=0)return 0;
    return (ax*by>ay*bx);
    }
void solve(ll l,ll r,ll x,ll fa){//dfs分治
    ll poss=l;
    for(ll i=l+1;i<=r;i++)
    if(p[i].x<p[poss].x||(p[i].x==p[poss].x&&p[i].y<p[poss].y))poss=i;
    swap(p[l],p[poss]);
    plk=p[l];
    Ans[p[l].id]=x;
    sort(p+l+1,p+r+1,cmp);//极角排序
    ll now=l+1;
    for(ll i=head[x];i;i=edge[i].nxt){
        ll to=edge[i].to;
        if(to==fa)continue;
        solve(now,now+size[to]-1,to,x);//分配
        now+=size[to];
        }
    }
int main(){
    scanf("%lld",&n);
    for(ll i=1;i<n;i++){
        ll a,b;
        scanf("%lld%lld",&a,&b);
        add_edge(a,b);
        add_edge(b,a);
        }
    for(ll i=1;i<=n;i++){
        ll x,y;
        scanf("%lld%lld",&x,&y);
        p[i]=(sett){x,y,i};
        }
    aux(1,0);
    solve(1,n,1,0);
    for(ll i=1;i<=n;i++)
    printf("%lld ",Ans[i]);
}