大致题意：
有一个人想要获得p个经验点和q元钱。现在给出n份工作，每份工作每天能得到Ai的经验值和Bi的钱，问最少需要工作多少天，
能使得总经验值>=p，总钱>=q。
先对给出的n份工作以及原点O(0,0)，x轴的最大点(maxx,0)，y轴最大点(0,maxy)构建凸包，根据凸组合，可知凸包上所有得点以
及凸包边上的点都可以由一天时间得到，那么只要求出射线O~P(p,q)与凸包的交点，即可求出最后的结果。
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 模型转换
原文地址：https://www.cnblogs.com/cjbiantai/p/9370155.html
时间： 2024-09-28 23:02:28