计算几何-多边形内核判定-HPI-poj3335

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

先解决一个问题，什么是多边形内核

形象的说就是在多边形中可以找到一个区域安放一台360°摄像头，能够监视到整个凸多边形区域

用手在多边形内侧摸一圈，凹凸不平？！

对，就是这个感觉。

借助数学必修5线性规划的思想，可以将多边形的n条边

看做n个线性约束条件

然后，在二位笛卡尔坐标系下求交集就好了（显然）

证明！

题目：poj3335

好，是不是直接上半平面交？！

我想给一组数据，poj上的，反正吓得我一弹，冷静的加了一行

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

先看数据

#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||fabs(Cross(l.v,A-l.P))<eps;}

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=100+10;
D P[maxn];L l[maxn];
int main(){
    int T,n,cnt;
    for(scanf("%d",&T);T--;){
        scanf("%d",&n),cnt=0;        
        for(int i=1;i<=n;i++){
            db a,b;scanf("%lf%lf",&a,&b);
            P[i]=D(a,b);
        }
      for(int i=n;i>=2;i--)
          l[++cnt]=L(P[i],P[i-1]-P[i]);
      l[++cnt]=L(P[1],P[n]-P[1]);
      printf("%s
",HPI(l,n,P)>=3?"YES":"NO");
    }
  return 0;
}