POJ-3111 K Best---二分求最大化平均值

题目链接：

https://cn.vjudge.net/problem/POJ-3111

题目大意：

卖宝救夫：Demy要卖珠宝，n件分别价值v_i 重 w_i，她希望保留k件使得

最大。

解题思路：

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
using namespace std;
const int INF = 1e7;
const int maxn = 100005;
int n, m;
double w[maxn], v[maxn], y[maxn];
bool judge(double x)
{
    for(int i = 1; i <= n; i++)
        y[i] = v[i] - x * w[i];
    sort(y + 1, y + n + 1);

    double sum = 0;
    for(int i = 1, j = n; i <= m; i++, j--)
        sum += y[j];
    return sum >= 0;
}
struct node
{
    double x;
    int id;
    bool operator <(const node& a)const
    {
        return x > a.x;
    }
}ans[maxn];
int main()
{
    cin >> n >> m;
    for(int i = 1; i <= n; i++)
        cin >> v[i] >> w[i];
    double l = 0, r = INF;
    for(int i = 0; i < 50; i++)
    {
        double mid = (l + r) / 2;
        if(judge(mid))l = mid;
        else r = mid;
    }
    for(int i = 1; i <= n; i++)
    {
        ans[i].id = i;
        ans[i].x = v[i] - l * w[i];
    }
    sort(ans + 1, ans + n + 1);
    printf("%d", ans[1].id);
    for(int i = 2; i <= m; i++)
    {
        printf(" %d", ans[i].id);
    }
    puts("");
    return 0;
}