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-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 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)==1||(sign(y)==0&&sign(x)==-1);}
    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==-1) return 1; else if (a==1) return 0; else return cmp(y,k1.y)==-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 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[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];}
    line push(db eps){//向左手边平移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<point> convexHull(vector<point>ps){
    int n = ps.size();if(n<=1)return ps;
    sort(ps.begin(),ps.end());
    vector<point> qs(n*2);int k=0;
    for(int i=0;i<n;qs[k++]=ps[i++])
        while (k>1&&cross(qs[k-1]-qs[k-2],ps[i]-qs[k-2])<=0)--k;
    for(int i=n-2,t=k;i>=0;qs[k++]=ps[i--])
        while (k>t&&cross(qs[k-1]-qs[k-2],ps[i]-qs[k-2])<=0)--k;
    qs.resize(k-1);
    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=0;i<m;i++){
        db mx = 0;int id=-1;
        for(int j=0;j<m;j++){
            db tmp = p[j].dis(proj(p[i],p[(i+1)%m],p[j]));
            if(cmp(tmp,mx)>0){
                mx=tmp,id=j;
            }
        }
        mp[make_pair(i,(i+1)%m)]=id;
    }
}
point tmp;
int main(){
    scanf("%d",&t);
    for(int cas=1;cas<=t;cas++){
        scanf("%d",&n);
        for(int i=0;i<n;i++){
            scanf("%lf%lf",&tmp.x,&tmp.y);
            p.push_back(tmp);
        }
        p=convexHull(p);
        if(p.size()<3){
            printf("Swarm %d Parallelogram Area: 0.0000
",cas);
            continue;
        }
        init();
        db ans = 1e15;
        int m = p.size();
        for(int i=0;i<m;i++){
            for(int j=i+1;j<m;j++) {
                ans = min(ans,check(i, (i + 1) % m, j,(j+1)%m));
            }
        }
        printf("Swarm %d Parallelogram Area: %.4f
",cas,ans);
        p.clear();mp.clear();
    }
}
/**
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