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-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);}
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)==1||(sign(y)==0&&sign(x)==-1);}
};
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))>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);
}
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);}
    point dir(){ return p[1]-p[0];}
};
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){
    sort(L.begin(),L.end());deque<line> q;
    for(int i=0;i<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 t,n;
point p[1551];
bool cw(){//时针
    db s=0;
    for(int i=1;i<n-1;i++){
        s+=cross(p[i]-p[0],p[i+1]-p[0]);
    }
    return s>0;
}
vector<line>l;
int main(){
    scanf("%d",&t);
    while (t--){
        scanf("%d",&n);
        for(int i=0;i<n;i++){
            scanf("%lf%lf",&p[i].x,&p[i].y);
        }
        if(!cw())reverse(p,p+n);
        for(int i=0;i<n;i++){
            l.push_back(line(p[i],p[(i+1)%n]));
        }
        l=getHL(l);
        if(l.size()<3){
            printf("0.00\n");
        } else{
            vector<point> a;
            for(int i=0;i<l.size();i++){
                a.push_back(getLL(l[i],l[(i+1)%l.size()]));
            }
            db ans = 0;
            for(int i=1;i<a.size()-1;i++)
                ans+=cross(a[i]-a[0],a[i+1]-a[0]);
            ans/=2;
            printf("%.2f\n",ans);
        }
        l.clear();
    }
}