sgu278：Fuel(线性规划)

题目大意：
$~~~~~~$对于$n$个元素，每个元素有三个非负整数属性$a,b,c$，总共有两个正整数限制$A,B$，求一个非负实数解集$X$，使得$\Sigma a_ix_i \leq A,\Sigma b_ix_i \leq B$且$\Sigma c_ix_i$最大。

分析：
$~~~~~~$我们很明显可以把三变量$a,b,c$化成两变量$d=\frac{a}{c},e=\frac{b}{c}$$($也就是取一个共同的单位$1)$，以及$y$表示$cx$。
$~~~~~~$$\Sigma a_ix_i \leq A \Rightarrow \Sigma d_iy_i \leq A$，$\Sigma b_ix_i \leq B \Rightarrow \Sigma e_iy_i \leq B$。
$~~~~~~$因为$x$是实数，所以我们的最优解要么只取一个元素，要么取的元素满足$\Sigma d_iy_i = A,\Sigma e_iy_i = B$。
$~~~~~~$将每个元素考虑成平面上一个点$(d,e)$，表示一个单位该元素消耗为$(d,e)$。对于平面上任选的点集，$(\Sigma d_iz_i,\Sigma e_iz_i),\Sigma z_i=1$就是这些点集构成图包内的所有点。如果要满足$\Sigma d_iy_i = A,\Sigma e_iy_i = B$，这个点集构成的凸包与线段$t=(A,B)$的交集就是组合成一个单位消耗$(a,b),\frac{a}{b}=\frac{A}{B}(a\leq A,b\leq B)$的新元素的所有解$g = (a,b)$，定义$l(p)$为线段$p$的长度，那么$l(t)/l(g)$即为这个解对应的答案。
$~~~~~~$因为$l(t)$已经确定，我们要让$l(g)$尽量小，那么只能是线段与凸包的交点处，因此我们求出整个凸包与线段$t$的离原点近的那个交点即可。

AC code：

#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <algorithm>
#include <string>
#include <sstream>
#include <iostream>
#include <map>
#include <set>
#include <list>
#include <stack>
#include <queue>
#include <vector>
#define pb push_back
#define mp make_pair
#define sqr(x) ((x)*(x))
typedef long long LL;
typedef double DB;
typedef long double LD;
using namespace std;

const int MAXN = 75009;
const DB eps = 1e-9;

int n, A, B;
struct pot
{
    DB x, y;
    pot() {x = y = 0;}
    pot(DB _x, DB _y):x(_x),y(_y){}

    friend pot operator + (const pot &a, const pot &b) {return pot(a.x+b.x, a.y+b.y);}
    friend pot operator - (const pot &a, const pot &b) {return pot(a.x-b.x, a.y-b.y);}
    friend pot operator * (const pot &a, DB k) {return pot(a.x*k, a.y*k);}
    friend pot operator / (const pot &a, DB k) {return pot(a.x/k, a.y/k);}
    friend DB operator * (const pot &a, const pot &b) {return a.x*b.x+a.y*b.y;}
    friend DB operator ^ (const pot &a, const pot &b) {return a.x*b.y-a.y*b.x;}
    friend bool operator == (const pot &a, const pot &b) {return fabs(a.x-b.x) <= eps && fabs(a.y-b.y) <= eps;}
    friend bool operator < (const pot &a, const pot &b)
    {
        if(fabs(a.x-b.x) <= eps) return a.y < b.y;
        else return a.x < b.x;
    }
    DB size() {return sqrt(sqr(x)+sqr(y));}

}node[MAXN], o, t;

pot st[MAXN];
int top;

DB ans;

bool cross(const pot &a, const pot &b, const pot &c, const pot &d)
{
    return ((c-a)^(b-a))*((d-a)^(b-a)) <= 0 && ((b-c)^(d-c))*((a-c)^(d-c)) <= 0;
}

pot get(const pot &a, const pot &b, const pot &c, const pot &d)
{
    pot p = b-a, q = d-c, w = a-c;
    DB k = (w^q)/(q^p);
    return a+p*k;
}

int main()
{
    #ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
    #endif

    scanf("%d%d%d", &n, &A, &B);
    o = pot(0, 0), t = pot(A, B);
    for(int i = 1; i <= n; ++i)
    {
        int a, b, c;
        scanf("%d%d%d", &a, &b, &c);
        node[i] = pot((DB)a/c, (DB)b/c);
        ans = max(ans, min(A/node[i].x, B/node[i].y));
    }
    sort(node+1, node+n+1);
    st[++top] = node[1], st[++top] = node[2];
    for(int i = 3; i <= n; ++i)
    {
        if(node[i] == node[i-1]) continue;
        while(top >= 2 && ((st[top]-node[i])^(st[top-1]-node[i])) < 0) top--;
        st[++top] = node[i];
    }
    int j = top;
    for(int i = n-1; i >= 1; --i)
    {
        if(node[i] == node[i-1]) continue;
        while(top-1 >= j && ((st[top]-node[i])^(st[top-1]-node[i])) < 0) top--;
        st[++top] = node[i];
    }
    st[0] = st[top];

    for(int i = 0; i < top; ++i)
        if(cross(st[i], st[i+1], o, t))
            ans = max(ans, t.size()/get(st[i], st[i+1], o, t).size());

    printf("%.6lf\n", ans);

    #ifndef ONLINE_JUDGE
    fclose(stdin);
    fclose(stdout);
    #endif
    return 0;
}