【包含内容】

0/1分数规划

 

一般形式:   然后选择做多k 个或 至少k 个 ,  这样的数 删去类的 使得 结果 R  最大 , 最优的问题.

我们先定义一个函数F(L):= ，显然这只是对目标式的一个简单的变形。分离参数，得到F(L):=sigma((。这时我们就会发现，如果L已知的话，a[i]-L*b[i]就是已知的，当然x[i]是未知的。记d[i]=a[i]-L*b[i]，那么F(L):=

 

一般采用 二分答案,  L  检索 F[l]  与  0  的关系 来判断  L 是否是最优

F(L)=  即

也就是说，如果一个方案使得F(L)>0   说明了这组方案可以得到一个比现在的L更优的一个L，

既然有一个更优的解，那么为什么不用呢？

F(L) <  0  说明  此时 答案过大 .

 

例如  Poj 2976

http://poj.org/problem?id=2976

选择 k 个删除使得的答案最大 ,  采用二分答案. 判优

#include <iostream>
#include <stdio.h>
#include <math.h>
#include <algorithm>
#include <string.h>
//#include <bits/stdc++.h>
#define rep(i,a,n) for(int i=a;i<n;i++)
#define per(i,a,n) for(int i=n-1;i>=a;i--)
typedef long long ll;
const int maxn = 1e5+10;
using namespace std;
const double esp = 1e-5;


int n,k;


double a[maxn],b[maxn],s[maxn];
int main()
{
    while(~scanf("%d %d",&n,&k))
    {
        if(n==0&&k==0) break;
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));
        memset(s,0,sizeof(s));
        rep(i,1,n+1)
            scanf("%lf",&a[i]);
        rep(i,1,n+1)
            scanf("%lf",&b[i]);
        double l = 0.0;
        double r = 1000000.0;

        while( l+esp < r)
        {
            double mid = (l+r)*1.0/2;
            rep(i,1,n+1)
                s[i] = a[i]-mid*b[i];
            sort(s+1,s+n+1);
            double sum =0;
            rep(i,k+1,n+1) sum+=s[i];
            if( sum >=0)
                l = mid;
            else
                r = mid;
        }
        printf("%d\n",(int)(l*100+0.5));
    }
    return 0;
}

例2 :牛客网暑期ACM多校训练营（第五场）
A

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

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 

备注:

1≤ n≤ 105

0≤ k < n

1≤ s[i],c[i] ≤ 103

选择 至多K 个 删除, 那么 排序后 从后面选择 n-k 个 , 然后 依次比较前k 个是否选择

#include <iostream>
#include <math.h>
#include <bits/stdc++.h>
#define rep(i,a,n) for(int i=a;i<n;i++)
#define per(i,a,n) for(int i=n-1;i>=a;i--)
typedef long long ll;
const int maxn = 1e5+10;
using namespace std;
const double esp = 1e-5;


int n,k;
double a[maxn];
double b[maxn];
double s[maxn];
bool judge(double mid)
{
    rep(i,1,n+1)
        s[i] = a[i]-mid*b[i];
    sort(s+1,s+n+1);
    double sum = 0;
    rep(i,k+1,n+1)
        sum+=s[i];
    double mmx = sum;
    rep(i,1,k+1)
        mmx = max(sum+s[i],mmx);

    if(mmx>=0)
        return true;
    return false;
}
int main()
{
    scanf("%d %d",&n,&k);
    rep(i,1,n+1)
        scanf("%lf",&b[i]);
    rep(i,1,n+1)
        scanf("%lf",&a[i]);
    double l =0.0 , r = 1.0;
    rep(i,1,n+1)
    {
        a[i] = a[i]*b[i];
        r = max(a[i],r);
    }
    while(l +esp <r)
    {
        double mid = (l+r)*1.0/2;
        if(judge(mid))
            l = mid;
        else
            r = mid;
       // cout<<l<<" "<<r<<endl;
    }
    //cout<<l<<" "<<r<<endl;
    printf("%.10f\n",l);
    return 0;
}