C. Freelancer's Dreams

Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

http://www.codeforces.com/contest/605/problem/C

Description

Mikhail the Freelancer dreams of two things: to become a cool programmer and to buy a flat in Moscow. To become a cool programmer, he needs at least p experience points, and a desired flat in Moscow costs q dollars. Mikhail is determined to follow his dreams and registered at a freelance site.

He has suggestions to work on n distinct projects. Mikhail has already evaluated that the participation in the i-th project will increase his experience by ai per day and bring bi dollars per day. As freelance work implies flexible working hours, Mikhail is free to stop working on one project at any time and start working on another project. Doing so, he receives the respective share of experience and money. Mikhail is only trying to become a cool programmer, so he is able to work only on one project at any moment of time.

Find the real value, equal to the minimum number of days Mikhail needs to make his dream come true.

For example, suppose Mikhail is suggested to work on three projects and a1 = 6, b1 = 2, a2 = 1, b2 = 3, a3 = 2, b3 = 6. Also, p = 20and q = 20. In order to achieve his aims Mikhail has to work for 2.5 days on both first and third projects. Indeed,a1·2.5 + a2·0 + a3·2.5 = 6·2.5 + 1·0 + 2·2.5 = 20 and b1·2.5 + b2·0 + b3·2.5 = 2·2.5 + 3·0 + 6·2.5 = 20.

Input

The first line of the input contains three integers np and q (1 ≤ n ≤ 100 000, 1 ≤ p, q ≤ 1 000 000) — the number of projects and the required number of experience and money.

Each of the next n lines contains two integers ai and bi (1 ≤ ai, bi ≤ 1 000 000) — the daily increase in experience and daily income for working on the i-th project.

Output

Print a real value — the minimum number of days Mikhail needs to get the required amount of experience and money. Your answer will be considered correct if its absolute or relative error does not exceed 10 - 6.

Namely: let's assume that your answer is a, and the answer of the jury is b. The checker program will consider your answer correct, if 

Sample Input

3 20 20
6 2
1 3
2 6

Sample Output

5.000000000000000

HINT

题意

给你一个2*n的矩阵A,然后让你找到一个n*2的矩阵B

要求使得矩阵中的所有数的和尽量小,使得A*B = [X,Y]

要求X>=P,Y>=Q

输出最小和

题解:

转化成计算几何问题,相当于给了你n个向量,然后你需要一个线性组合使得某个向量,超过P,Q这个点

我们可以将所有向量组成一个凸包。

这里需要一个证明,假设B矩阵的和是1的话,所有向量的线性组合最多达到这个凸包的边界上。

如果n=1的话,很显然成立,n=2也显然成立,n=3必然比n=2指的更加短,所以这个结论就成立了?

然后我们就二分答案就好了,然后再判一判(P,Q)这个点是否在这个凸包内部就好了咯。

代码:

#include<iostream>
#include<math.h>
#include<algorithm>
#include<cstring>
#include<cstdio> using namespace std; #define maxn 100005
const double EP = 1e-;
const int MAXV = ;
const double PI = 3.14159265; /* 基本几何结构 */
struct POINT
{
double x;
double y;
POINT(double a=, double b=) { x=a; y=b;} //constructor
};
POINT operator - (POINT A,POINT B){return POINT(A.x-B.x,A.y-B.y);}
struct LINESEG
{
POINT s;
POINT e;
LINESEG(POINT a, POINT b) { s=a; e=b;}
LINESEG() { }
};
double multiply(POINT sp,POINT ep,POINT op)
{
return((sp.x-op.x)*(ep.y-op.y)-(ep.x-op.x)*(sp.y-op.y));
}
bool InsideConvexPolygon(int vcount,POINT polygon[],POINT q) // 可用于三角形!
{
POINT p;
LINESEG l;
int i;
p.x=;p.y=;
for(i=;i<vcount;i++) // 寻找一个肯定在多边形polygon内的点p:多边形顶点平均值
{
p.x+=polygon[i].x;
p.y+=polygon[i].y;
}
p.x /= vcount;
p.y /= vcount; for(i=;i<vcount;i++)
{
l.s=polygon[i];l.e=polygon[(i+)%vcount];
if(multiply(p,l.e,l.s)*multiply(q,l.e,l.s)<) /* 点p和点q在边l的两侧,说明点q肯定在多边形外 */
break;
}
return (i==vcount);
}
double Cross(POINT a,POINT b)
{
return a.x*b.y-a.y*b.x;
}
bool cmp1(POINT a,POINT b)
{
if(fabs(a.x-b.x)<EP)
return a.y<b.y;
return a.x<b.x;
}
int CH(POINT* p,int n,POINT* ch)
{
sort(p,p+n,cmp1);
int m=;
for(int i=;i<n;i++)
{
while(m>&&Cross(ch[m-]-ch[m-],p[i]-ch[m-])<=)m--;
ch[m++]=p[i];
}
int k=m;
for(int i=n-;i>=;i--)
{
while(m>k&&Cross(ch[m-]-ch[m-],p[i]-ch[m-])<=)m--;
ch[m++]=p[i];
}
if(n>)m--;
return m;
}
int N,tot;
POINT fin,pp[maxn],t[maxn],T[maxn];
int check(double x)
{
for(int i=;i<tot;i++)
T[i].x = t[i].x*x,T[i].y = t[i].y*x;
if(InsideConvexPolygon(tot,T,fin))return ;
return ;
}
int main()
{
memset(pp,,sizeof(pp));
memset(t,,sizeof(t));
memset(T,,sizeof(T));
double X=,Y=;
scanf("%d",&N);cin>>fin.x>>fin.y;
for(int i=;i<N;i++)
{
double x,y;
scanf("%lf%lf",&x,&y);
X = max(X,x);
Y = max(Y,y);
pp[i].x = x,pp[i].y = y;
}
pp[N].x = ,pp[N].y = ;
pp[N+].x = X,pp[N+].y = ;
pp[N+].x = ,pp[N+].y = Y;
N+=;
tot = CH(pp,N,t);
double l = ,r = 999990009.0;
for(int i=;i<=;i++)
{
double mid = (l+r)/2.0;
if(check(mid))r=mid;
else l=mid;
}
printf("%.15f\n",l);
}

Codeforces Round #335 (Div. 1) C. Freelancer's Dreams 计算几何的更多相关文章

  1. Codeforces Round #335 (Div. 1)--C. Freelancer's Dreams 线性规划对偶问题+三分

    题意:p, q,都是整数. sigma(Ai * ki)>= p, sigma(Bi * ki) >= q; ans = sigma(ki).输出ans的最小值 约束条件2个,但是变量k有 ...

  2. Codeforces Round #335 (Div. 2) B. Testing Robots 水题

    B. Testing Robots Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://www.codeforces.com/contest/606 ...

  3. Codeforces Round #335 (Div. 2) D. Lazy Student 构造

    D. Lazy Student Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/606/probl ...

  4. Codeforces Round #335 (Div. 2) C. Sorting Railway Cars 动态规划

    C. Sorting Railway Cars Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://www.codeforces.com/conte ...

  5. Codeforces Round #335 (Div. 2) A. Magic Spheres 水题

    A. Magic Spheres Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://www.codeforces.com/contest/606/ ...

  6. Codeforces Round #335 (Div. 2) D. Lazy Student 贪心+构造

    题目链接: http://codeforces.com/contest/606/problem/D D. Lazy Student time limit per test2 secondsmemory ...

  7. Codeforces Round #335 (Div. 2)

    水 A - Magic Spheres 这题也卡了很久很久,关键是“至少”,所以只要判断多出来的是否比需要的多就行了. #include <bits/stdc++.h> using nam ...

  8. Codeforces Round #335 (Div. 2) A. Magic Spheres 模拟

    A. Magic Spheres   Carl is a beginner magician. He has a blue, b violet and c orange magic spheres. ...

  9. Codeforces Round #335 (Div. 2) D. Lazy Student 贪心

    D. Lazy Student   Student Vladislav came to his programming exam completely unprepared as usual. He ...

随机推荐

  1. linux中ulimit作用

    一.作用 Linux对于每个用户,系统限制其最大进程数.为提高性能,可以根据设备资源情况,设置各linux 用户的最大进程数. ulimit主要是用来限制进程对资源的使用情况的,它支持各种类型的限制, ...

  2. 【C++】非原创|统计代码覆盖率(一:C++)

    也是转别人的,因为我c++好菜好菜啊... http://blog.chinaunix.net/uid-23741326-id-3316943.html c++跟C基本是一样的,统计覆盖率,需要生成g ...

  3. Golang连接Oracle数据库

    Golang连接Oracle的库有很多,比较常见的如下: 不过,oralce 只提供了 oci8 的接口,必须通过它来调用,所以下面方案都逃不过相关设置. 1.go-db-oracle 地址: htt ...

  4. bjfu1262 优先队列

    比较典型的应用优先队列的题.题目是在一个长为n的数组中,依次问m个数中的最小值.那么把值和下标做成一个结构体,放进优先队列里,每次移动窗口就把该T的T掉,剩下的最小值就是答案,复杂度nlogn,轻松a ...

  5. 两个队列+k叉哈夫曼树 HDU 5884

    // 两个队列+k叉哈夫曼树 HDU 5884 // camp题解: // 题意:nn个有序序列的归并排序.每次可以选择不超过kk个序列进行合并,合并代价为这些序列的长度和.总的合并代价不能超过TT, ...

  6. 牛课--C/C++

    引用是除指针外另一个可以产生多态效果的手段. //引用是除指针外另一个可以产生多态效果的手段. #include<iostream> using namespace std; class ...

  7. Spring MVC + Spriing + MyBatis整合,写给新人

    开发环境: 开发工具:MyEclipse 8.6 数据库:MySQL 操作系统:WIN8.1 Jar包: Spirng和SpringMVC版本:3.2.9 MyBatis版本:3.2.8 其他关联Ja ...

  8. Oracle 游标使用(转)

    这个文档几乎包含了oracle游标使用的方方面面,全部通过了测试 ; ; dbms_output.put_line(sql) loop dbms_output.put_line( ; ; ; r_te ...

  9. oracle max()函数和min()函数

    当需要了解一列中的最大值时,可以使用MAX()函数:同样,当需要了解一列中的最小值时,可以使用MIN()函数.语法如下. SELECT          MAX (column_name) / MIN ...

  10. C# JackLib系列之GdiHelper圆角矩形的快速生成

    using System; using System.Collections.Generic; using System.Linq; using System.Text; using System.D ...