01分数规划牛客网 入门题

链接：https://www.nowcoder.com/acm/contest/143/A
来源：牛客网

题目描述
Kanade selected n courses in the university. The academic credit of the i-th course is s[i] and the score of the i-th course is c[i].

At the university where she attended, the final score of her is

Now she can delete at most k courses and she want to know what the highest final score that can get.
输入描述:

The first line has two positive integers n,k

The second line has n positive integers s[i]

The third line has n positive integers c[i]

输出描述:

Output the highest final score, your answer is correct if and only if the absolute error with the standard answer is no more than 10-5

示例1
输入
复制

3 1
1 2 3
3 2 1

输出
复制

2.33333333333

说明

Delete the third course and the final score is frac{2*2+3*1}{2+1}=frac{7}{3}

备注:

1≤ n≤ 105

0≤ k < n

1≤ s[i],c[i] ≤ 103

题意：给你两个数组， s[i] , c[i] ，再给你一个式子时累加求和的相比求最大值的式子，同时你最多可以再其中删去 K 个字母，问最终最大收益是多少？

思路分析 ： 经典的 01分数规划问题，最优的结果一定是去掉越多的值越优，想一想就知道为什么了

代码示例 ：

#include <bits/stdc++.h>
using namespace std;
#define ll long long
const int maxn = 1e6+5;
const int mod = 1e9+7;
const double eps = 1e-9;
const double pi = acos(-1.0);
const int inf = 0x3f3f3f3f;

int n, k;
double s[maxn], c[maxn];
double a[maxn];

bool check(double x){
    for(int i = 1; i <= n; i++){
        a[i] = (c[i]-x)*s[i];
    }    
    sort(a+1, a+1+n);
    
    double sum=0;
    for(int i = k+1; i <= n; i++){
        sum += a[i];
    }
    //printf("*** %lf
", sum);
    if (sum > 0) return true;
    return false; 
}

int main() {
    //freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
    cin >> n >> k;
    
    for(int i = 1; i <= n; i++) {
        scanf("%lf", &s[i]);
    }    
    for(int i = 1; i <= n; i++) {
        scanf("%lf", &c[i]);
    }
    double l = 0, r = 1000.0;
    for(int i = 1; i <= 50; i++){
        double mid = (l+r)/2;
        if (check(mid)) l = mid;
        else r = mid;
    }
    printf("%.11lf
", l);
    return 0;
}