POJ2976: Dropping Tests 题解-优快云博客

经典的01分数规划问题
01分数规划通常解决这样的问题
给定 $n$ 个元素，每个元素都有两个属性值 $a_i$ , $b_i$ ，要选出其中的k个 $c_1,c_2...c_k$ ，最大化 $\frac{\sum_{i=1}^{k}a_{c_i}}{\sum_{i=1}^kb_{c_i}}$
解决这样的问题我们可以考虑二分答案
假设当前二分出的答案是 $x$ ，我们考虑是否能选出k个元素满足条件
可以发现

\frac{\sum_{i = 1}^{k} a_{c_{i}}}{\sum_{i = 1}^{k} b_{c_{i}}} \geq x ⟺ \sum_{i = 1}^{k} a_{c_{i}} - x \sum_{i = 1}^{k} b_{c_{i}} \geq 0 ⟺ \sum_{i = 1}^{k} (a_{c_{i}} - x b_{c_{i}}) \geq 0

$\frac{\sum_{i=1}^{k}a_{c_i}}{\sum_{i=1}^{k}b_{c_i}}\geq x \Longleftrightarrow \sum_{i=1}^{k}a_{c_i}-x\sum_{i=1}^{k}b_{c_i}\geq 0 \Longleftrightarrow \sum_{i=1}^{k}(a_{c_i}-xb_{c_i})\geq 0$
所以将所有元素按照

ai−xbiai−xbi $a_i-xb_i$ 排序，取前

kk $k$ 个判

s u m

$sum$ 是否大于0就好(这题中是前

n−kn−k $n-k$ 个)

#include <cstdio>
#include <iostream>
#include <cstring>
#include <string>
#include <cstdlib>
#include <utility>
#include <cctype>
#include <algorithm>
#include <bitset>
#include <set>
#include <map>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <cmath>
#define LL long long
#define LB long double
#define x first
#define y second
#define Pair pair<int,int>
#define pb push_back
#define pf push_front
#define mp make_pair
#define LOWBIT(x) x & (-x)
#define DEBUG(...)
using namespace std;

const int MOD=1e9+7;
const LL LINF=2e16;
const int INF=2e9;
const int magic=348;
const double eps=1e-10;

inline int getint()
{
    char ch;int res;bool f;
    while (!isdigit(ch=getchar()) && ch!='-') {}
    if (ch=='-') f=false,res=0; else f=true,res=ch-'0';
    while (isdigit(ch=getchar())) res=res*10+ch-'0';
    return f?res:-res;
}

int n,k;
double l,r,mid;

struct node
{
    double a,b;
    inline bool operator < (const node &x) const {return a-mid*b>x.a-mid*x.b;}
}a[10048];

inline bool check()
{
    int i;
    sort(a+1,a+n+1);
    double A=0,B=0;
    for (i=1;i<=n-k;i++) A+=a[i].a,B+=a[i].b;
    if (A/B>=mid) return true; else return false;
}

int main ()
{
    int i;double ans;
    while (scanf("%d%d",&n,&k) && !(!n && !k))
    {
        for (i=1;i<=n;i++) a[i].a=getint();
        for (i=1;i<=n;i++) a[i].b=getint();
        l=0;r=1;ans=0;
        while (r-l>1e-6)
        {
            mid=(l+r)/2;
            if (check()) ans=mid,l=mid; else r=mid;
        }
        printf("%d\n",int(ans*100+0.5));
    }
    return 0;
}