poj1279

板子题，求多边形内核面积。

话说jls的板子返回的是边，然后我就在冥思苦想怎么根据割边求面积啊。。

然后发现自己果然是个傻逼，求一下交点存起来就好了。。。

//板子题到此为止了

 #include <iostream>

 #include <cstdio>

 #include <algorithm>

 #include <vector>

 #include <cmath>

 #include <deque>

 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);}

 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};}

     db getP()const { return sign(y)==||(sign(y)==&&sign(x)==-);}

 };

 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));}

 int compareangle(point k1,point k2){

     return k1.getP()<k2.getP()||(k1.getP()==k2.getP()&&sign(cross(k1,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);

 }

 struct line{

     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[])>);}

     point dir(){ return p[]-p[];}

 };

 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){

     sort(L.begin(),L.end());deque<line> q;

     for(int i=;i<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);

         for(int i=;i<n;i++){

             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("0.00\n");

         } else{

             vector<point> a;

             for(int i=;i<l.size();i++){

                 a.push_back(getLL(l[i],l[(i+)%l.size()]));

             }

             db ans = ;

             for(int i=;i<a.size()-;i++)

                 ans+=cross(a[i]-a[],a[i+]-a[]);

             ans/=;

             printf("%.2f\n",ans);

         }

         l.clear();

     }

 }