CF605C. Freelancer’s Dreams

题目描述

Solution

实际上就是给定ai,bi,A,Ba_i,b_i,A,Bai,bi,A,B，求n维向量(x1..xn)(x1..x_n)(x1..xn)，使得：
∑i=1naixi≥A∑i=1nbixi≥Bminz=∑ixi\sum_{i=1}^na_ix_i\geq A\\ \sum_{i=1}^nb_ix_i\geq B\\ min\;z=\sum_i{x_i} i=1∑naixi≥Ai=1∑nbixi≥Bminz=i∑xi
答案就是minzmin\;zminz
这是一个简单线性规划问题，转化为对偶形式，可得：
∀i=1..naiy1+biy2<=1maxz=Ay1+By2y1,y2≥0\forall_{i=1..n}a_iy_1+b_iy_2<=1\\ max\;z={Ay_1+By_2}\\ y_1,y_2\geq 0 ∀i=1..naiy1+biy2<=1maxz=Ay1+By2y1,y2≥0
可以把上面的1式转化为：
∀i=1..ny1<=1−biy2ai\forall_{i=1..n}y_1<=\frac{1-b_iy_2}{a_i} ∀i=1..ny1<=ai1−biy2

现在的变量是y1,y2y_1,y_2y1,y2，相当于求二维平面的半平面交，可以求一个凸包解决。

但显然有更简单的方法：我们确定了y2y_2y2之后可以唯一确定y1y_1y1，并且该情况下答案显然是一个凸函数，所以可以直接三分y2y_2y2去求出极值点。

时间复杂度O(nlgn)O(nlgn)O(nlgn)。

#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <ctime>
#include <cassert>
#include <string.h>
//#include <unordered_set>
//#include <unordered_map>
//#include <bits/stdc++.h>#define MP(A,B) make_pair(A,B)
#define PB(A) push_back(A)
#define SIZE(A) ((int)A.size())
#define LEN(A) ((int)A.length())
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define fi first
#define se secondusing namespace std;template<typename T>inline bool upmin(T &x,T y) { return y<x?x=y,1:0; }
template<typename T>inline bool upmax(T &x,T y) { return x<y?x=y,1:0; }typedef long long ll;
typedef unsigned long long ull;
typedef long double lod;
typedef pair<int,int> PR;
typedef vector<int> VI;const lod eps=1e-12;
const lod pi=acos(-1);
const int oo=1<<30;
const ll loo=1ll<<62;
const int mods=998244353;
const int MAXN=600005;
const int INF=0x3f3f3f3f;//1061109567
/*--------------------------------------------------------------------*/
inline int read()
{int f=1,x=0; char c=getchar();while (c<'0'||c>'9') { if (c=='-') f=-1; c=getchar(); }while (c>='0'&&c<='9') { x=(x<<3)+(x<<1)+(c^48); c=getchar(); }return x*f;
}
int n,A,B,a[MAXN],b[MAXN];
lod check(lod y2)
{lod y1=1e6;for (int i=1;i<=n;i++) upmin(y1,(1-y2*b[i])/a[i]);if (y1<eps) return -1;return y1*A+y2*B;
}
int main()
{n=read(),A=read(),B=read();for (int i=1;i<=n;i++) a[i]=read(),b[i]=read();lod l=0,r=1e6;while (r-l>eps){lod midl=l+(r-l)/3,midr=r-(r-l)/3;if (check(midl)+eps<check(midr)) l=midl;else r=midr;}printf("%.11lf\n",(double)check(l));return 0;
}