半平面交&多边形内核。因为没注意了点的情况自闭了。

https://blog.csdn.net/qq_40861916/article/details/83541403

这个说的贼好。

多边形内核就是多边形内部的一块区域/一个点,能看到多边形的任何地方。

怎么求呢。

首先每条边要逆时针。

然后我们对所有的边按照与 +x轴的逆时针夹角进行排序。从小到大。

这之后我们每次用双端队列维护已经求出的多边形。

每加入一条新边的话,我们check一下

这个图说的太好了!话说这个图的blog我在上面放了链接不算盗用叭。。。

然后就没了。。

 #include <iostream>
#include <cmath>
#include <algorithm>
#include <cstdio>
#include <vector>
#include <deque>
#define rep(x) for(int i=0;i<x;i++)
using namespace std;
typedef double db;
const db eps=1e-;
const db pi=acos(-);
int sign(db k){
if (k>eps) return ; else if (k<-eps) return -; return ;
}
int cmp(db k1,db k2){return sign(k1-k2);}
int inmid(db k1,db k2,db k3){return sign(k1-k3)*sign(k2-k3)<=;}// k3 在 [k1,k2] 内
struct point{
db x,y;
point operator + (const point &k1) const{return (point){k1.x+x,k1.y+y};}
point operator - (const point &k1) const{return (point){x-k1.x,y-k1.y};}
point operator * (db k1) const{return (point){x*k1,y*k1};}
point operator / (db k1) const{return (point){x/k1,y/k1};}
int operator == (const point &k1) const{return cmp(x,k1.x)==&&cmp(y,k1.y)==;}
// 逆时针旋转
point turn(db k1){return (point){x*cos(k1)-y*sin(k1),x*sin(k1)+y*cos(k1)};}
point turn90(){return (point){-y,x};}
bool operator < (const point k1) const{
int a=cmp(x,k1.x);
if (a==-) return ; else if (a==) return ; else return cmp(y,k1.y)==-;
}
db abs(){return sqrt(x*x+y*y);}
db abs2(){return x*x+y*y;}
db dis(point k1){return ((*this)-k1).abs();}
point unit(){db w=abs(); return (point){x/w,y/w};}
void scan(){double k1,k2; scanf("%lf%lf",&k1,&k2); x=k1; y=k2;}
void print(){printf("%.11lf %.11lf\n",x,y);}
db getw(){return atan2(y,x);}
point getdel(){if (sign(x)==-||(sign(x)==&&sign(y)==-)) return (*this)*(-); else return (*this);}
int getP() const{return sign(y)==||(sign(y)==&&sign(x)==-);}
};
int inmid(point k1,point k2,point k3){return inmid(k1.x,k2.x,k3.x)&&inmid(k1.y,k2.y,k3.y);}
db cross(point k1,point k2){return k1.x*k2.y-k1.y*k2.x;}
db dot(point k1,point k2){return k1.x*k2.x+k1.y*k2.y;}
db rad(point k1,point k2){return atan2(cross(k1,k2),dot(k1,k2));}
// -pi -> pi
int compareangle (point k1,point k2){
return k1.getP()<k2.getP()||(k1.getP()==k2.getP()&&sign(cross(k1,k2))>);
}
point proj(point k1,point k2,point q){ // q 到直线 k1,k2 的投影
point k=k2-k1; return k1+k*(dot(q-k1,k)/k.abs2());
}
point reflect(point k1,point k2,point q){return proj(k1,k2,q)*-q;}
int clockwise(point k1,point k2,point k3){// k1 k2 k3 逆时针 1 顺时针 -1 否则 0
return sign(cross(k2-k1,k3-k1));
}
int checkLL(point k1,point k2,point k3,point k4){// 求直线 (L) 线段 (S)k1,k2 和 k3,k4 的交点
return cmp(cross(k3-k1,k4-k1),cross(k3-k2,k4-k2))!=;
}
point getLL(point k1,point k2,point k3,point k4){
db w1=cross(k1-k3,k4-k3),w2=cross(k4-k3,k2-k3); return (k1*w2+k2*w1)/(w1+w2);
}
int intersect(db l1,db r1,db l2,db r2){
if (l1>r1) swap(l1,r1); if (l2>r2) swap(l2,r2); return cmp(r1,l2)!=-&&cmp(r2,l1)!=-;
}
int checkSS(point k1,point k2,point k3,point k4){
return intersect(k1.x,k2.x,k3.x,k4.x)&&intersect(k1.y,k2.y,k3.y,k4.y)&&
sign(cross(k3-k1,k4-k1))*sign(cross(k3-k2,k4-k2))<=&&
sign(cross(k1-k3,k2-k3))*sign(cross(k1-k4,k2-k4))<=;
}
db disSP(point k1,point k2,point q){
point k3=proj(k1,k2,q);
if (inmid(k1,k2,k3)) return q.dis(k3); else return min(q.dis(k1),q.dis(k2));
}
db disSS(point k1,point k2,point k3,point k4){
if (checkSS(k1,k2,k3,k4)) return ;
else return min(min(disSP(k1,k2,k3),disSP(k1,k2,k4)),min(disSP(k3,k4,k1),disSP(k3,k4,k2)));
}
int onS(point k1,point k2,point q){return inmid(k1,k2,q)&&sign(cross(k1-q,k2-k1))==;}
struct circle{
point o; db r;
void scan(){o.scan(); scanf("%lf",&r);}
int inside(point k){return cmp(r,o.dis(k));}
};
struct line{
// p[0]->p[1]
point p[];
line(point k1,point k2){p[]=k1; p[]=k2;}
point& operator [] (int k){return p[k];}
int include(point k){return sign(cross(p[]-p[],k-p[]))>=;}//非严格包含。
point dir(){return p[]-p[];}
line push(){ // 向外 ( 左手边 ) 平移 eps
const db eps = 1e-;
point delta=(p[]-p[]).turn90().unit()*eps;
return {p[]-delta,p[]-delta};
}
};
point getLL(line k1,line k2){return getLL(k1[],k1[],k2[],k2[]);}
int parallel(line k1,line k2){return sign(cross(k1.dir(),k2.dir()))==;}
int sameDir(line k1,line k2){return parallel(k1,k2)&&sign(dot(k1.dir(),k2.dir()))==;}
int operator < (line k1,line k2){
if (sameDir(k1,k2)) return k2.include(k1[]);
return compareangle(k1.dir(),k2.dir());
}
int checkpos(line k1,line k2,line k3){return k3.include(getLL(k1,k2));}
vector<line> getHL(vector<line> &L){ // 求半平面交 , 半平面是逆时针方向 , 输出按照逆时针
sort(L.begin(),L.end()); deque<line> q;
for (int i=;i<(int)L.size();i++){
if (i&&sameDir(L[i],L[i-])) continue;
while (q.size()>&&!checkpos(q[q.size()-],q[q.size()-],L[i])) q.pop_back();
while (q.size()>&&!checkpos(q[],q[],L[i])) q.pop_front();
q.push_back(L[i]);
}
while (q.size()>&&!checkpos(q[q.size()-],q[q.size()-],q[])) q.pop_back();
while (q.size()>&&!checkpos(q[],q[],q[q.size()-])) q.pop_front();
vector<line>ans; for (int i=;i<q.size();i++) ans.push_back(q[i]);
return ans;
}
int t,n;
point p[];
bool cw(){//时针
db s=;
for(int i=;i<n-;i++){
s+=cross(p[i]-p[],p[i+]-p[]);
}
return s>;
}
vector<line> l;
int main(){
scanf("%d",&t);
while (t--){
scanf("%d",&n);
rep(n) scanf("%lf%lf",&p[i].x,&p[i].y);
if(!cw())reverse(p,p+n);
for(int i=;i<n;i++){
l.push_back(line(p[i],p[(i+)%n]));
}
l=getHL(l);
if(l.size()>=){
printf("YES\n");
} else{
printf("NO\n");
}
l.clear();
}
}
/**
1
17
2 -1 2 -2 1 -2 0 -1 -1 -2 -2 -2 -2 -1 -1 0 -2 1 -2 2 -1 2 0 1 1 2 3 2 3 1 2 1 1 0 */

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