APIO2018 Circle selection 选圆圈

APIO2018 Circle selection 选圆圈

题意:

题目传送门

题解:

似乎网上题解都是KDTree啊……
反正似乎裸的KDTree,稍微旋转一下角度,似乎就不会被卡到\(n^2\)了……不过如果\(1e9\)\(double\)平方一下会爆精,开个\(long \ \ double\)苟过去……

Code:

#include <bits/stdc++.h>
using namespace std;
const int N = 3e5 + 50;
#define double long double 
const double alpha = 1.785;
const double eps = 1e-4;

int n, nw;
struct Circle {
  double v[2], r;
  int idx;
  void Read() {
    double x, y;
    scanf("%Lf%Lf%Lf", &x, &y, &r);
    v[0] = x * cos(alpha) + y * sin(alpha);
    v[1] = y * cos(alpha) - x * sin(alpha);
  }
  friend bool operator < (Circle a, Circle b) {
    return a.v[nw] < b.v[nw];
  }
}p1[N], p2[N], Nw;
int ret[N];

double Sqr(double x) { return x * x; }

int Check(Circle a, Circle b) {
  return Sqr(a.r + b.r) - Sqr(a.v[0] - b.v[0]) - Sqr(a.v[1] - b.v[1]) >= -eps;
}

bool Cmp(Circle a, Circle b) {
  return a.r == b.r ? (a.idx < b.idx) : (a.r > b.r);
}

namespace KDT {
  int ls[N], rs[N];
  double up[N], dn[N], le[N], ri[N];
  int rt;
  void Merge(int x, int y) {
    le[x] = min(le[x], le[y]);
    ri[x] = max(ri[x], ri[y]);
    dn[x] = min(dn[x], dn[y]);
    up[x] = max(up[x], up[y]);
  }
  void Update(int o) {
    le[o] = p1[o].v[0] - p1[o].r; 
    ri[o] = p1[o].v[0] + p1[o].r; 
    dn[o] = p1[o].v[1] - p1[o].r; 
    up[o] = p1[o].v[1] + p1[o].r; 
    if(ls[o]) Merge(o, ls[o]);
    if(rs[o]) Merge(o, rs[o]);
  }
  int Build(int l, int r, int d) {
    if(l > r) return 0;
    if(l == r) return (Update(l), l);
    int mid = (l + r) >> 1; nw = d & 1;
    nth_element(p1 + l, p1 + mid, p1 + r + 1);
    ls[mid] = Build(l, mid - 1, d + 1);
    rs[mid] = Build(mid + 1, r, d + 1);
    Update(mid);
    return mid;
  }
  int In(int o) {
    return le[o] - Nw.v[0] - Nw.r <= eps && Nw.v[0] - Nw.r - ri[o] <= eps && dn[o] - Nw.v[1] - Nw.r <= eps && Nw.v[1] - Nw.r - up[o] <= eps; 
  }
  void Dfs(int o) {
    if(!In(o)) return ;
    if(!ret[p1[o].idx] && Check(Nw, p1[o])) ret[p1[o].idx] = Nw.idx;
    if(ls[o]) Dfs(ls[o]);
    if(rs[o]) Dfs(rs[o]);
  }
}

int main() {
  scanf("%d", &n);
  for(int i = 1; i <= n; i++) p1[i].Read(), p1[i].idx = i, p2[i] = p1[i];
  sort(p2 + 1, p2 + 1 + n, Cmp);
  KDT::rt = KDT::Build(1, n, 0);
  for(int i = 1; i <= n; i++) {
    if(!ret[p2[i].idx]) {
      Nw = p2[i];
      KDT::Dfs(KDT::rt);
      ret[p2[i].idx] = p2[i].idx;
    }
  }
  for(int i = 1; i <= n; i++) printf("%d ", ret[i]);
  puts("");
  return 0;
}
posted @ 2019-04-01 20:24  Apocrypha  阅读(253)  评论(0编辑  收藏  举报