计算几何-poj2451-HPI

Posted 2020-09-06 Jason-Cow

This article is made by Jason-Cow.
Welcome to reprint.
But please post the article\'s address.

 

题意，求半平面交，并且保证有解

poj2451

 

给张图边界（|x,y|<=1000,题目为1e4）

 

别问我为什么TLE

我当成多组处理没有判断文末符号，加上就！A！啦！

直接上模板

 

  1 #include <algorithm>
  2 #include <iostream>
  3 #include <cstring>
  4 #include <cstdlib>
  5 #include <cstdio>
  6 #include <vector>
  7 #include <cmath>
  8 #include <queue>
  9 #include <map>
 10 #include <set>
 11 using namespace std;
 12 #define sqr(x) ((x)*(x))
 13 #define RG register
 14 #define op operator
 15 #define IL inline
 16 typedef double db;
 17 typedef bool bl;
 18 const db pi=acos(-1.0),eps=1e-10;
 19 struct D{
 20   db x,y;
 21   D(db x=0.0,db y=0.0):x(x),y(y){}
 22 };
 23 typedef D V;
 24 bl operator<(D A,D B){return A.x<B.x||(A.x==B.x&&A.y<B.y);}
 25 V operator+(V A,V B){return V(A.x+B.x,A.y+B.y);}
 26 V operator-(V A,V B){return V(A.x-B.x,A.y-B.y);}
 27 V operator*(V A,db N){return V(A.x*N,A.y*N);}
 28 V operator/(V A,db N){return V(A.x/N,A.y/N);}
 29 
 30 db Ang(db x){return(x*180.0/pi);}
 31 db Rad(db x){return(x*pi/180.0);}
 32 V Rotate(V A,db a){return V(A.x*cos(a)-A.y*sin(a),A.x*sin(a)+A.y*cos(a));}
 33 db Dis(D A,D B){return sqrt(sqr(A.x-B.x)+sqr(A.y-B.y));}
 34 db Cross(V A,V B){return A.x*B.y-A.y*B.x;}
 35 
 36 db Area(D*R,int n){
 37     db S=0.0;
 38     for(int i=1;i<n;i++)S+=Cross(R[i]-R[1],R[i+1]-R[1]);
 39     return S/2;
 40 }
 41 
 42 db Length(D*R,int n){
 43     db C=0.0;
 44     for(int i=2;i<=n;i++)C+=Dis(R[i],R[i-1]);
 45     return C+Dis(R[n],R[1]);
 46 }
 47 
 48 struct L{
 49   D P,v;
 50   db a;
 51   L(){}
 52   L(D P,V v):P(P),v(v){a=atan2(v.y,v.x);}
 53   bool operator<(const L x)const{return a<x.a;}
 54 };
 55 
 56 D Intersect(L a,L b){
 57   V u=a.P-b.P;
 58   return a.P+a.v*(Cross(b.v,u)/Cross(a.v,b.v));
 59 }
 60 
 61 bool Left(L l,D A){
 62   return Cross(l.v,A-l.P)>0;
 63 }
 64 
 65 int HPI(L*l,int n,D*ans){
 66   int head,tail,m=0;
 67   D*P=new D[n];L*q=new L[n];
 68   sort(l+1,l+n+1),q[head=tail=0]=l[1];
 69   for(int i=2;i<=n;i++){
 70     while(head<tail && !Left(l[i],P[tail-1]))tail--;
 71     while(head<tail && !Left(l[i],P[head]))  head++;
 72     q[++tail]=l[i];
 73     if(fabs(Cross(q[tail].v,q[tail-1].v))<eps){
 74       tail--;
 75       if(Left(q[tail],l[i].P))q[tail]=l[i];
 76     }
 77     if(head<tail)P[tail-1]=Intersect(q[tail-1],q[tail]);
 78   }
 79   while(head<tail && !Left(q[head],P[tail-1]))tail--;
 80   if(tail-head<=1)return 0;
 81   P[tail]=Intersect(q[tail],q[head]);
 82   for(int i=head;i<=tail;i++)ans[++m]=P[i];
 83   return m;
 84 }
 85 
 86 const int maxn=2e4+10;
 87 L l[maxn];D A[maxn];
 88 int main(){
 89   D LU(0,1e4),LD(0,0),RD(1e4,0),RU(1e4,1e4);
 90   for(int n;scanf("%d",&n)!=EOF;){
 91     for(int i=1;i<=n;i++){
 92       db a,b,c,d;
 93       scanf("%lf%lf%lf%lf",&a,&b,&c,&d);
 94       D A(a,b),B(c,d);
 95       l[i]=L(A,B-A);
 96     }
 97     l[++n]=L(LU,V(0,-1));
 98     l[++n]=L(LD,V(1,0));
 99     l[++n]=L(RD,V(0,1));
100     l[++n]=L(RU,V(-1,0));
101     int m=HPI(l,n,A);
102     printf("%.1lf\\n",Area(A,m));
103   }
104   return 0;
105 }