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
1 #include <iostream> 2 #include <cstdio> 3 #include <algorithm> 4 #include <vector> 5 #include <cmath> 6 #include <deque> 7 8 using namespace std; 9 typedef double db; 10 const db eps=1e-6; 11 const db pi=acos(-1); 12 int sign(db k){ 13 if (k>eps) return 1; else if (k<-eps) return -1; return 0; 14 } 15 int cmp(db k1,db k2){return sign(k1-k2);} 16 struct point{ 17 db x,y; 18 point operator + (const point &k1) const{return (point){k1.x+x,k1.y+y};} 19 point operator - (const point &k1) const{return (point){x-k1.x,y-k1.y};} 20 point operator * (db k1) const{return (point){x*k1,y*k1};} 21 point operator / (db k1) const{return (point){x/k1,y/k1};} 22 db getP()const { return sign(y)==1||(sign(y)==0&&sign(x)==-1);} 23 }; 24 db cross(point k1,point k2){ return k1.x*k2.y-k1.y*k2.x;} 25 db dot(point k1,point k2){ return k1.x*k2.x+k1.y*k2.y;} 26 db rad(point k1,point k2){ return atan2(cross(k1,k2),dot(k1,k2));} 27 int compareangle(point k1,point k2){ 28 return k1.getP()<k2.getP()||(k1.getP()==k2.getP()&&sign(cross(k1,k2))>0); 29 } 30 point getLL(point k1,point k2,point k3,point k4){ 31 db w1=cross(k1-k3,k4-k3),w2=cross(k4-k3,k2-k3); 32 return (k1*w2+k2*w1)/(w1+w2); 33 } 34 struct line{ 35 point p[2]; 36 line(point k1,point k2){p[0]=k1;p[1]=k2;} 37 point &operator[](int k){ return p[k];} 38 int include(point k){ return sign(cross(p[1]-p[0],k-p[0])>0);} 39 point dir(){ return p[1]-p[0];} 40 }; 41 point getLL(line k1,line k2){ 42 return getLL(k1[0],k1[1],k2[0],k2[1]); 43 } 44 int parallel(line k1,line k2){ return sign(cross(k1.dir(),k2.dir()))==0;} 45 int sameDir(line k1,line k2){ 46 return parallel(k1,k2)&&sign(dot(k1.dir(),k2.dir()))==1; 47 } 48 int operator <(line k1,line k2){ 49 if(sameDir(k1,k2))return k2.include(k1[0]); 50 return compareangle(k1.dir(),k2.dir()); 51 } 52 int checkpos(line k1,line k2,line k3){ return k3.include(getLL(k1,k2));} 53 vector<line> getHL(vector<line> &L){ 54 sort(L.begin(),L.end());deque<line> q; 55 for(int i=0;i<L.size();i++){ 56 if(i&&sameDir(L[i],L[i-1]))continue; 57 while (q.size()>1&&!checkpos(q[q.size()-2],q[q.size()-1],L[i]))q.pop_back(); 58 while (q.size()>1&&!checkpos(q[1],q[0],L[i]))q.pop_front(); 59 q.push_back(L[i]); 60 } 61 while (q.size()>2&&!checkpos(q[q.size()-2],q[q.size()-1],q[0]))q.pop_back(); 62 while (q.size()>2&&!checkpos(q[1],q[0],q[q.size()-1]))q.pop_front(); 63 vector<line> ans;for(int i=0;i<q.size();i++)ans.push_back(q[i]); 64 return ans; 65 } 66 int t,n; 67 point p[1551]; 68 bool cw(){//时针 69 db s=0; 70 for(int i=1;i<n-1;i++){ 71 s+=cross(p[i]-p[0],p[i+1]-p[0]); 72 } 73 return s>0; 74 } 75 vector<line>l; 76 int main(){ 77 scanf("%d",&t); 78 while (t--){ 79 scanf("%d",&n); 80 for(int i=0;i<n;i++){ 81 scanf("%lf%lf",&p[i].x,&p[i].y); 82 } 83 if(!cw())reverse(p,p+n); 84 for(int i=0;i<n;i++){ 85 l.push_back(line(p[i],p[(i+1)%n])); 86 } 87 l=getHL(l); 88 if(l.size()<3){ 89 printf("0.00 "); 90 } else{ 91 vector<point> a; 92 for(int i=0;i<l.size();i++){ 93 a.push_back(getLL(l[i],l[(i+1)%l.size()])); 94 } 95 db ans = 0; 96 for(int i=1;i<a.size()-1;i++) 97 ans+=cross(a[i]-a[0],a[i+1]-a[0]); 98 ans/=2; 99 printf("%.2f ",ans); 100 } 101 l.clear(); 102 } 103 }