大致題意：

　　　　有一個人想要獲得p個經驗點和q元錢。現在給出n份工作，每份工作每天能得到Ai的經驗值和Bi的錢，問最少需要工作多少天，

　　　　能使得總經驗值>=p，總錢>=q。

　　　　先對給出的n份工作以及原點O(0,0)，x軸的最大點(maxx,0)，y軸最大點(0,maxy)構建凸包，根據凸組合，可知凸包上所有得點以

　　　　及凸包邊上的點都可以由一天時間得到，那麽只要求出射線O~P(p,q)與凸包的交點，即可求出最後的結果。

  1 #include<cstdio>
  2 #include<iostream>
  3 #include<cstring>
  4
 
 #include<algorithm>
  5 #include<queue>
  6 #include<set>
  7 #include<map>
  8 #include<stack>
  9 #include<time.h>
 10 #include<cstdlib>
 11 #include<cmath>
 12 #include<list>
 13 using namespace std;
 14 #define MAXN 200100
 15 #define eps 1e-5
 16
 
 #define For(i,a,b) for(int i=a;i<=b;i++)
 17 #define Fore(i,a,b) for(int i=a;i>=b;i--)
 18 #define lson l,mid,rt<<1
 19 #define rson mid+1,r,rt<<1|1
 20 #define mkp make_pair
 21 #define pb push_back
 22 #define cr clear()
 23 #define sz size()
 24 #define met(a,b) memset(a,b,sizeof(a))
 25
 
 #define iossy ios::sync_with_stdio(false)
 26 #define fre freopen
 27 #define pi acos(-1.0)
 28 #define inf 1e9+9
 29 #define Vector Point
 30 const int Mod=1e9+7;
 31 typedef unsigned long long ull;
 32 typedef long long ll;
 33 typedef pair<int,int> pii;
 34 typedef pair<ll,ll> pll;
 35 int dcmp(double x){
 36     if(fabs(x)<=eps) return 0;
 37     return x<0?-1:1;
 38 }
 39 struct Point {
 40     double x,y;
 41     int id;
 42     int pre,nxt;
 43     Point(double x=0,double y=0) : x(x),y(y) {}
 44     Point operator - (const Point &a)const{ return Point(x-a.x,y-a.y); }
 45     Point operator + (const Point &a)const{ return Point(x+a.x,y+a.y); }
 46     Point operator * (const double &a)const{ return Point(x*a,y*a); }
 47     Point operator / (const double &a)const{ return Point(x/a,y/a); }
 48     bool operator < (const Point &a)const{    if(x==a.x) return y<a.y;return x<a.x; }
 49     bool operator == (const Point &a)const{ return dcmp(x-a.x)==0 && dcmp(y-a.y)==0; }
 50     void read(int iid=0) { scanf("%lf%lf",&x,&y);id=iid; }
 51     void out(){cout<<"Bug: "<<x<<" "<<y<<endl;}
 52 };
 53 inline double Cross(Vector a,Vector b) { return a.x*b.y-a.y*b.x; }
 54 inline double Dot(Vector a,Vector b) { return a.x*b.x+a.y*b.y; }
 55 inline double dis(Vector a) { return sqrt(Dot(a,a)); }
 56 int ConvexHull(Point *p,int n,Point *ch){
 57     int m=0;
 58     sort(p,p+n);
 59     For(i,0,n-1){
 60         p[i].id=i;
 61         while(m>1 && Cross(ch[m-1]-ch[m-2],p[i]-ch[m-2])<=0) m--;
 62         ch[m++]=p[i];
 63     }
 64     int k=m;
 65     Fore(i,n-2,0){
 66         while(m>k && Cross(ch[m-1]-ch[m-2],p[i]-ch[m-2])<=0) m--;
 67         ch[m++]=p[i];
 68     }
 69     if(n>1) m--;
 70     return m;
 71 }
 72 inline bool Onsegment(Point a,Point b1,Point b2){
 73     return dcmp(Cross(b1-a,b2-a))==0 && dcmp(Dot(b1-a,b2-a))<0;    
 74 }
 75 bool Intersect_Segm_Ray(Point a1,Vector u,Point b1,Point b2){
 76     double c1=Cross(b1-a1,u),c2=Cross(b2-a1,u);
 77     if(dcmp(c2)*dcmp(c1)>0) return 0;
 78     if(dcmp(c2)*dcmp(c1)<0){
 79         c1=Dot(b1-a1,u);c2=Dot(b2-a1,u);
 80         return dcmp(c1)>=0 && dcmp(c2)>=0;
 81     }
 82     if(c1==0) return dcmp(Dot(b1-a1,u))>0;
 83     if(c2==0) return dcmp(Dot(b2-a1,u))>0;
 84 }
 85 Point Intersect_Line_Point(Point p,Vector u,Point q,Vector v){
 86     Vector w=p-q;
 87     double t=Cross(v,w)/Cross(u,v);
 88     return p+u*t;
 89 }
 90 int n;
 91 double maxx,maxy;
 92 Point p[MAXN],cp;
 93 Point ch[MAXN];
 94 int solve(){
 95     double ans=inf;maxx=0;maxy=0;
 96     cin>>n;cp.read();
 97     For(i,0,n-1) p[i].read(),maxx=max(p[i].x,maxx),maxy=max(p[i].y,maxy);
 98     p[n]=Point(0,0);
 99     p[n+1]=Point(0,maxy);
100     p[n+2]=Point(maxx,0);
101     int m=ConvexHull(p,n+3,ch);
102     For(i,0,m-1){
103         if(ch[i]==Point(0,0) || ch[(i+1)%m]==Point(0,0)) continue;
104         if(Intersect_Segm_Ray(Point(0,0),cp,ch[(i+1)%m],ch[i])){
105             Point st=Intersect_Line_Point(cp,cp,ch[i],ch[(i+1)%m]-ch[i]);
106             ans=min(ans,dis(cp)/dis(st));
107         }
108     }
109     printf("%.15lf\n",ans);
110 }
111 int main(){
112 //  fre("in.txt","r",stdin);
113     int t=1;
114     while(t--)solve();
115     return 0;
116 }

[CodeForces-606E] Freelancer's Dreams 凸包 模型轉換