计算几何（二维数点问题）

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题解：题目比较明显的提示了二维数点，关键在于坐标的变换的推导，感觉是正常高中水平就可以推出，所以不在题解里赘述了（其实是画图太麻烦，懒的画）

#include<bits/stdc++.h>
using namespace std;
#define rep(i,a,b) for(int i=(a);i<b;i++)
#define pb push_back
#define db double
#define ll long long
#define fi first
#define se second
#define all(x) x.begin(),x.end()

const db pi = acos(-1.0);
const int N = 1e5+5;
double eps = 1e-8;
vector<pair<db,db> > point;
struct Node{
  db fi,se;
  int id;
};
vector<Node> pt;
vector<db> V; 
int res[N];
int C[N];
int lowbit(int x){return x&(-x);}
int query(int x){
  int t = 0;
  for(;x;x -= lowbit(x))t += C[x];
  return t;
}
void mofidy(int x,int d){
  for(;x<N;x+=lowbit(x))C[x]+=d;
  return;
}

int discrete(db x){
  return lower_bound(all(V),x) - V.begin() + 1;
}

int main(){
  int n,k;
  scanf("%d%d",&n,&k);
  point.resize(n);
  pt.resize(n);
  rep(i,0,n)scanf("%lf%lf",&point[i].fi,&point[i].se);
  while(k--){
    V.clear();
    memset(C,0,sizeof C);
    db l,r;scanf("%lf%lf",&l,&r);
    db sl = sinl(l/180.0*pi), cl = cosl(l/180.0*pi);
    db sr = sinl(r/180.0*pi), cr = cosl(r/180.0*pi);
    rep(i,0,n){
      db & x = point[i].fi;
      db & y = point[i].se;
      pt[i].fi = x * sr + y * cr;
      pt[i].se = x * sl + y * cl;
      pt[i].id = i;
      V.pb(pt[i].se);
    }
    sort(all(V));
    V.erase(unique(all(V),[&](db a,db b){return fabs(a-b) <= eps;}),V.end());
    sort(all(pt),[&](Node a ,Node b){
      return fabs(a.fi-b.fi)>eps?a.fi<b.fi:a.se>b.se;
    });
    rep(i,0,n){
      int yy = discrete(pt[i].second-eps);
      res[pt[i].id] += query(N-1) - query(yy-1);
      mofidy(yy,1);
    }
  }
  rep(i,0,n){
    printf("%d ",res[i]);
  }
  printf("\n");
}

标签：pt,int,res,rep,数点,yy,二维,几何,query
来源： https://www.cnblogs.com/mafengfa/p/16219411.html