0/1分数规划

学习了lyd书上的0/1分数规划，发现这类题目都有一个特点，就是求$\frac{\sum_{a_{i}*x_{i}}}{\sum_{b_{i}*x_{i}}}$的最大或者最小，再加一些限制取不取的条件.

POJ2976

二分答案+sort取前（n-k+1)个。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
using namespace std;
const int maxn = 1e3 + 50;
int a[maxn], b[maxn];
double diff[maxn];
int n, k;
int check(double x)
{
    for(int i = 1; i <= n; i++)
    {
        diff[i] = (double)a[i] - x * b[i];
    }
    sort(diff + 1, diff + n + 1);
    double ans = 0;
    for(int i = n; i >= k + 1; i--)
    {
        ans += diff[i];
    }
    if(ans >= 0.0 || fabs(ans) <= (1e-8)) return 1;
    else return 0;
}
int main()
{
    while(scanf("%d %d", &n, &k) != EOF && (n || k))
    {
        for(int i = 1; i <= n; i++) scanf("%d", &a[i]);
        for(int i = 1; i <= n; i++) scanf("%d", &b[i]);
        double ans = 0;
        double l = 0, r = 1;
        int T = 100;
        while(T--)
        {
            double mid = (l + r) / 2.0;
            if(check(mid))
            {
              //  printf("%f\n", mid);
                ans = mid;
                l = mid;
            }
            else
            {
                r = mid;
            }
        }
        printf("%.f\n", 100.0 * ans);
    }
    return 0;
}

Code

原文地址：https://www.cnblogs.com/wangwangyu/p/9735685.html