计算几何BZOJ2618凸包的交HPI

This article is made by Jason-Cow.
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bzoj2618

1A，看懂了题目就是读入的复杂度

把每个凸包看成m个约束条件

一起读入然后做一次HPI

O((nm)log(nm)) (nm<=500,时间完全可以承受)

/**************************************************************
    Problem: 2618
    User: Jaosn_liu
    Language: C++
    Result: Accepted
    Time:0 ms
    Memory:1332 kb
****************************************************************/

ACcode

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <vector>
#include <cmath>
#include <queue>
#include <map>
#include <set>
using namespace std;
#define sqr(x) ((x)*(x))
#define RG register
#define op operator
#define IL inline
typedef double db;
typedef bool bl;
const db pi=acos(-1.0),eps=1e-10;
struct D{
  db x,y;
  D(db x=0.0,db y=0.0):x(x),y(y){}
};
typedef D V;
bl operator<(D A,D B){return A.x<B.x||(A.x==B.x&&A.y<B.y);}
V operator+(V A,V B){return V(A.x+B.x,A.y+B.y);}
V operator-(V A,V B){return V(A.x-B.x,A.y-B.y);}
V operator*(V A,db N){return V(A.x*N,A.y*N);}
V operator/(V A,db N){return V(A.x/N,A.y/N);}

db Ang(db x){return(x*180.0/pi);}
db Rad(db x){return(x*pi/180.0);}
V Rotate(V A,db a){return V(A.x*cos(a)-A.y*sin(a),A.x*sin(a)+A.y*cos(a));}
db Dis(D A,D B){return sqrt(sqr(A.x-B.x)+sqr(A.y-B.y));}
db Cross(V A,V B){return A.x*B.y-A.y*B.x;}

db Area(D*R,int n){
  db S=0.0;
  for(int i=1;i<n;i++)S+=Cross(R[i]-R[1],R[i+1]-R[1]);
  return S/2;
}

db Length(D*R,int n){
  db C=0.0;
  for(int i=2;i<=n;i++)C+=Dis(R[i],R[i-1]);
  return C+Dis(R[n],R[1]);
}

struct L{
  D P,v;
  db a;
  L(){}
  L(D P,V v):P(P),v(v){a=atan2(v.y,v.x);}
  bool operator<(const L x)const{return a<x.a;}
};

D Intersect(L a,L b){
  V u=a.P-b.P;
  return a.P+a.v*(Cross(b.v,u)/Cross(a.v,b.v));
}

bool Left(L l,D A){
  return Cross(l.v,A-l.P)>0;
}

int HPI(L*l,int n,D*ans){
  int head,tail,m=0;
  D*P=new D[n];L*q=new L[n];
  sort(l+1,l+n+1),q[head=tail=0]=l[1];
  for(int i=2;i<=n;i++){
    while(head<tail && !Left(l[i],P[tail-1]))tail--;
    while(head<tail && !Left(l[i],P[head]))  head++;
    q[++tail]=l[i];
    if(fabs(Cross(q[tail].v,q[tail-1].v))<eps){
      tail--;
      if(Left(q[tail],l[i].P))q[tail]=l[i];
    }
    if(head<tail)P[tail-1]=Intersect(q[tail-1],q[tail]);
  }
  while(head<tail && !Left(q[head],P[tail-1]))tail--;
  if(tail-head<=1)return 0;
  P[tail]=Intersect(q[tail],q[head]);
  for(int i=head;i<=tail;i++)ans[++m]=P[i];
  return m;
}

const int maxn=500+10;
int n,m,cnt;
L l[maxn];
D A[maxn],Temp,Last,Now;
db a,b;

int main(){
    scanf("%d",&n);
    while(n--){
        scanf("%d",&m);
        scanf("%lf%lf",&a,&b);
        Last=D(a,b);
        Temp=Last;
        for(int i=2;i<=m;i++){
            scanf("%lf%lf",&a,&b);
            Now=D(a,b);
            l[++cnt]=L(Last,Now-Last);
            Last=Now;
        }
        l[++cnt]=L(Now,Temp-Now);
    }
    printf("%.3lf\n",Area(A,HPI(l,cnt,A)));
    return 0;
}