https://www.cnblogs.com/tellmewtf/p/4574438.html
这份代码不能AC，避雷，不过思路是对的，自己写就行了，还是很好写的。
样例二很容易画出来，然后思路就比较好想了。
唔，半平面交的tag也是很显然的东西。
现在考虑怎么做，其实一条直线只要在它一侧的点的个数满足限制，那么这个半平面就是部分合法的，我们就是要对所有合法半平面求交。
然后很容易想到扫描线的做法，那么这个题就做完了。
关于限制，在我代码的第145行，<=和==都能过。
这一点我也不太明白，理论上，哦草竟然没有三点共线，都过了才发现。。。
这样的话等于的时候其实就是小于等于，所以没什么影响。
其实扫描线是根本不怕三点共线这种东西的，，那这个题就挺简单的了，应该很难写错吧。
注意输出的时候用 .6f ，我也不懂为什么，反正我 .11f 就wa，好坑啊，，，先是被错误的题解坑了，又被输出坑。。。
#include <bits/stdc++.h>
using namespace std;
typedef double db;
const db eps=1e-6;
const db pi=acos(-1);
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 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)==0&&cmp(y,k1.y)==0;}
// 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==-1) return 1; else if (a==1) return 0; else return cmp(y,k1.y)==-1;
}
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)==-1||(sign(x)==0&&sign(y)==-1)) return (*this)*(-1); else return (*this);}
int getP() const{return sign(y)==1||(sign(y)==0&&sign(x)==-1);}
};
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){//极角排序+
return k1.getP()<k2.getP()||(k1.getP()==k2.getP()&&sign(cross(k1,k2))>0);
}
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)*2-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))!=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);
}
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)!=-1&&cmp(r2,l1)!=-1;
}
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))<=0&&
sign(cross(k1-k3,k2-k3))*sign(cross(k1-k4,k2-k4))<=0;
}
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 0;
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))==0;}
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[2];
line(){}
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;}
point dir(){return p[1]-p[0];}
line push(){ // 向外 ( 左手边 ) 平移 eps
const db eps = 1e-6;
point delta=(p[1]-p[0]).turn90().unit()*eps;
return {p[0]-delta,p[1]-delta};
}
};
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 operator < (line k1,line k2){
if (sameDir(k1,k2)) return k2.include(k1[0]);
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(auto x:L){
// x[0].print();
// x[1].print();
// printf("\n");
// }
for (int i=0;i<(int)L.size();i++){
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]);
return ans;
}
int checkLL(line k1,line k2){
return checkLL(k1.p[0],k1.p[1],k2.p[0],k2.p[1]);
}
db area(vector<point> A){ // 多边形用 vector<point> 表示 , 逆时针
db ans=0;
for (int i=0;i<A.size();i++)
ans+=cross(A[i],A[(i+1)%A.size()]);
return ans/2;
}
int T,n;
point p[1005];
vector<line> hp;
vector<point> v;
void slove(int id){
int bound = n/3-1;
v.clear();
for(int i=1;i<=n;i++)if(i!=id)v.push_back(p[i]-p[id]);
sort(v.begin(),v.end(),compareangle);
int m = v.size();
for(int i=0;i<m;i++)v.push_back(v[i]);
int t=0;
for(int l=0,r;l<m;l=r+1){
r=l;
point _180 = {-v[l].x,-v[l].y};
t=max(l,t);
while (t<l+m-1&&(sign(cross(v[t+1],_180)>=0&&cross(v[t+1],v[l])<=0))) t++;
if(m-(t-l+1)==bound){
hp.push_back(line(p[id],p[id]+v[l]));
// printf("%.11f %.11f %.11f %.11f\n",p[id].x,p[id].y,p[id].x+v[l].x,p[id].y+v[l].y);
}
while (sign(cross(v[l],v[r+1]))==0&&sign(dot(v[l],v[r+1]))==0)r++;
}
}
int main(){
scanf("%d",&T);int cas = 0;
while (T--){
hp.clear();
v.clear();
cas++;
scanf("%d",&n);
for(int i=1;i<=n;i++)scanf("%lf%lf",&p[i].x,&p[i].y);
for(int i=1;i<=n;i++){
slove(i);
}
hp = getHL(hp);
if(hp.size()<3){printf("Case #%d: 0\n",cas);continue;}
int m=hp.size();
v.clear();
for(int i=0;i<m;i++){
v.push_back(getLL(hp[i],hp[(i+1)%m]));
}
db s = area(v);
printf("Case #%d: %.6f\n",cas,s);
}
}