gym101657 C

嘻嘻嘻嘻，从读题到过题大概一个小时？

这套题题面太长了。。。就挑短的写。。

题意很简单。   给出平面上n个点，求一个面积最小的平行四边形覆盖这n个点。

显然要先求凸包。

然后枚举边就可以了。我一开始脑子抽了，只枚举了相邻的。。(今天下午四点多才吃上早饭很痛苦。感觉神智不清，昨晚补题补到3点。。。)

然后我们要 分别找出 离这两条边最远的 点吧，然后通过简单的平移操作求直线交点，再用叉积搞一下就阔以了。

虽然理论上 n^3不会t，但我还是t了。。。所以我们先预处理出来离每条边最远的点是哪一个。就阔以了。

记得特判一下没有凸包的情况，虽然不知道有没有这种数据。。。

 #include <bits/stdc++.h>
 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 abs(){return sqrt(x*x+y*y);}
     db dis(point k1){return ((*this)-k1).abs();}
     point turn90(){ return point{-y,x};}
     db getP()const { return sign(y)==||(sign(y)==&&sign(x)==-);}
     point unit(){db w=abs(); return point{x/w,y/w};}
     db abs2(){return x*x+y*y;}
     bool operator < (const point k1) const{
         int a=cmp(x,k1.x);
         if (a==-) return ; else if (a==) return ; else return cmp(y,k1.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));}
 int compareangle(point k1,point k2){
     return k1.getP()<k2.getP()||(k1.getP()==k2.getP()&&sign(cross(k1,k2))>);
 }
 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 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[];}
     line push(db eps){//向左手边平移eps
         //const db eps=1e-6;
         point delta=(p[]-p[]).turn90().unit()*eps;
         return {p[]-delta,p[]-delta};
     }
 };
 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<point> convexHull(vector<point>ps){
     int n = ps.size();if(n<=)return ps;
     sort(ps.begin(),ps.end());
     vector<point> qs(n*);int k=;
     for(int i=;i<n;qs[k++]=ps[i++])
         while (k>&&cross(qs[k-]-qs[k-],ps[i]-qs[k-])<=)--k;
     for(int i=n-,t=k;i>=;qs[k++]=ps[i--])
         while (k>t&&cross(qs[k-]-qs[k-],ps[i]-qs[k-])<=)--k;
     qs.resize(k-);
     return qs;
 }
 int t,n;
 vector<point> p;point s1,t1,s2,t2;
 map<pair<int,int>,int> mp;
 db check(int a,int b,int c,int d){
     int id1=mp[make_pair(a,b)],id2=mp[make_pair(c,d)];
     s1=p[id1],s2=p[id1]+p[a]-p[b];
     point t1 = getLL(p[c],p[d],s1,s2);//一个交点
     s1 = p[id2],s2=p[id2]+p[c]-p[d];
     point t2 = getLL(p[a],p[b],s1,s2);
     point bb = getLL(p[a],p[b],p[c],p[d]);
     db S = abs(cross(bb-t1,bb-t2));
     return S;
 }
 void init(){
     int m = p.size();
     for(int i=;i<m;i++){
         db mx = ;int id=-;
         for(int j=;j<m;j++){
             db tmp = p[j].dis(proj(p[i],p[(i+)%m],p[j]));
             if(cmp(tmp,mx)>){
                 mx=tmp,id=j;
             }
         }
         mp[make_pair(i,(i+)%m)]=id;
     }
 }
 point tmp;
 int main(){
     scanf("%d",&t);
     for(int cas=;cas<=t;cas++){
         scanf("%d",&n);
         for(int i=;i<n;i++){
             scanf("%lf%lf",&tmp.x,&tmp.y);
             p.push_back(tmp);
         }
         p=convexHull(p);
         if(p.size()<){
             printf("Swarm %d Parallelogram Area: 0.0000\n",cas);
             continue;
         }
         init();
         db ans = 1e15;
         int m = p.size();
         for(int i=;i<m;i++){
             for(int j=i+;j<m;j++) {
                 ans = min(ans,check(i, (i + ) % m, j,(j+)%m));
             }
         }
         printf("Swarm %d Parallelogram Area: %.4f\n",cas,ans);
         p.clear();mp.clear();
     }
 }
 /**
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 4
 0 0
 0 1
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  */