# zsbenn

### Blog Stats

• Posts - 919
• Stories - 2
• Comments - 7
• Trackbacks - 0

## 【模板】多边形的核（非严格的半平面角）——poj3335

   int include(point k){return sign(cross(p[1]-p[0],k-p[0]))>=0;}//k在l左端
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<vector>
#include<algorithm>
#include<queue>
using namespace std;
#define N 205

typedef double db;
const db eps=1e-6;
const db pi=acos(-1.0);
int sign(db k){
if (k>eps) return 1; else if (k<-eps) return -1; return 0;
}
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)<=0;}// k3 在 [k1,k2] 内
struct point{
db x,y;
point(){}
point(db x,db y):x(x),y(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 getP() const{return sign(y)==1||(sign(y)==0&&sign(x)>=0);}
};
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;}
int comp(point k1,point k2){
if(k1.getP()==k2.getP())return sign(cross(k1,k2))>0;
return k1.getP()<k2.getP();
}

struct line{
point p[2];
line(point k1,point k2){p[0]=k1; p[1]=k2;}
point& operator [] (int k){return p[k];}
int include(point k){return sign(cross(p[1]-p[0],k-p[0]))>=0;}//k在l左端
point dir(){return p[1]-p[0];}
};

//输入的点是顺时针：ans<0,逆时针：ans>0
bool judge(vector<point> v){
double ans=0;
for(int i=1;i<v.size()-1;i++)
ans+=cross(v[i]-v[0],v[i+1]-v[0]);
return ans>0;
}
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);
}
point getLL(line k1,line k2){return getLL(k1[0],k1[1],k2[0],k2[1]);}
int parallel(line k1,line k2){return sign(cross(k1.dir(),k2.dir()))==0;}
int sameDir(line k1,line k2){return parallel(k1,k2)&&sign(dot(k1.dir(),k2.dir()))==1;}
int checkpos(line k1,line k2,line k3){return k3.include(getLL(k1,k2));}//k1,k2交点在 k3 左端
int operator<(line k1,line k2){//按极角排序，角度相同的从左到右排
if(sameDir(k1,k2))return k2.include(k1[0]);
return comp(k1.dir(),k2.dir());
}
int 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=0;i<(int)L.size();i++){
//cout<<L[i][0].x<<" "<<L[i][0].y<<" "<<L[i][1].x<<" "<<L[i][1].y<<'\n';
if (i&&sameDir(L[i],L[i-1])) continue;
while (q.size()>1&&!checkpos(q[q.size()-2],q[q.size()-1],L[i])) q.pop_back();
while (q.size()>1&&!checkpos(q[1],q[0],L[i])) q.pop_front();
q.push_back(L[i]);
}
while (q.size()>2&&!checkpos(q[q.size()-2],q[q.size()-1],q[0])) q.pop_back();
while (q.size()>2&&!checkpos(q[1],q[0],q[q.size()-1])) q.pop_front();
vector<line>ans; for (int i=0;i<q.size();i++) ans.push_back(q[i]);
if(q.size()<3)return 0;
return 1;
}
vector<point>v;
vector<line>L;
int n;

int main(){
int t;cin>>t;
while(t--){
v.clear();L.clear();

scanf("%d",&n);
for(int i=1;i<=n;i++){
point t;
scanf("%lf%lf",&t.x,&t.y);
v.push_back(t);
}

if(judge(v)){//点逆时针
for(int i=0;i<n;i++)
L.push_back(line(v[i],v[(i+1)%n]));
}else {//点顺时针
for(int i=0;i<n;i++)
L.push_back(line(v[(i+1)%n],v[i]));
}

int res = getHL(L);
if(!res)puts("NO");
else puts("YES");
}
}

posted on 2020-02-27 20:48  zsben  阅读(...)  评论(...编辑  收藏