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;
}