01分数规划

问题

\(01\)分数规划是用来解决这样一类问题

有\(n\)个物品，每个物品都有一个属性\(p\)和\(w\)。要从中选出\(k\)个物品使得\(\frac{\sum\limits_{i=1}^kp_i}{\sum\limits_{i=1}^kw_i}\)最大，输出最大值。要求是个分数

思想

首先二分一个答案\(x\)。
然后将上面的问题转化为要选\(k\)个物品使得\[\frac{\sum\limits_{i=1}^kp_i}{\sum\limits_{i=1}^kw_i} \geq x\]
即\[\sum\limits_{i=1}^kp_i-\sum\limits_{i=1}^kw_i\times x \ge 0\]
即\[\sum\limits_{i=1}^k{p_i-w_i \times x} \ge 0\]
所以对于每个物品，按照上面这个式子排个序。看最大的k个是否满足条件即可。如果满足条件就统计出答案。

例题

51nod 1257

代码

/*
* @author: wxyww
* @date:   2019-02-09 08:30:09
* @last modified time: 2019-02-09 09:00:36
*/
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<iostream>
#include<cstdlib>
#include<cmath>
#include<ctime>
#include<bitset>
using namespace std;
typedef long long ll;
const int n = 50000 + 10;
const double eps = 1e-9;
ll read() {
    ll x=0,f=1;char c=getchar();
    while(c<'0'||c>'9') {
        if(c=='-') f=-1;
        c=getchar();
    }
    while(c>='0'&&c<='9') {
        x=x*10+c-'0';
        c=getchar();
    }
    return x*f;
}
struct node {
    int w,p;
    double x;
}a[n];
bool tmp(node x,node y) {
    return x.x > y.x;
}
int n,k;
int check(double w,int &x,int &y) {
    for(int i = 1;i <= n;++i)
        a[i].x = double(a[i].w) - double(a[i].p * w);
    sort(a + 1,a + n + 1,tmp);
    double ans = 0;
    for(int i = 1;i <= k;++i) ans += a[i].x;
    if(ans >= 0) {
        x = 0,y = 0;
        for(int i = 1;i <= k;++i) {
            x += a[i].w;
            y += a[i].p;
        }
        int g = __gcd(x,y);
        x /= g; y /= g;
        return 1;
    }
    return 0;
}
int main() {
    n = read();
    k = read();
    double r = 0;
    for(int i = 1;i <= n;++i) a[i].p = read(),a[i].w = read(),r = max(r,(double)a[i].w);
    double l = 0;
    int x,y;
    while(r - l > eps) {
        double mid = (l + r) / 2;
        if(check(mid,x,y)) l = mid;
        else r = mid;
    }
    printf("%d/%d\n",x,y);
    return 0;
}