目标函数转化求解Set partitioning问题

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描述

There are some positive integer numbers, and you want to divide them into two nonempty sets (every integer should be in and only in one set) and get a value descripted below:

Assume that set A has n1 numbers A1, A2, ...,An1; set B has n2 numbers B1, B2, ...Bn2.

Let

Then

We want to know the largest ans.

输入

The first line of the input is an integer T (T <= 20), which stands for the number of test cases you need to solve.

Every test case begins with an integer N (2 <= N <= 10000), then followed by N positive integers on the next line, these numbers will not exceed 1000.

输出

For every test case, you should output "Case #k: " first, where k indicates the case number and starts at 1. Then output the answer rounded to two digits after the decimal point. See sample for more details.

样例输入

2
3
1 1 1
3
1 2 3

样例输出

Case #1: 0.00
Case #2: 7.25

思路：

将输入数据从小到大排列后，问题为求一个位置i （0<=i<n），使得以i为界的两个部分（第i个位置的元素属于第一部分）满足目标函数：

max

如果线性搜索i ，计算每个ans求得最大值，遍历时间为O（n）,计算时间为O（n），总的时间复杂度为O（n2），在10000的数据规模下会超时。

此处，需要将目标函数进行拆分转义，得到化简后的目标函数为：

所有元素的平方和（定值）+ n1*v2*v2+n2*v1*v1-2*v1*v2*n ;

其中，n1= i+1,n2=n-i-1 . v1,v2也容易求得。

这样计算的时间复杂度降为O（1），总的时间复杂度为O（n）。

代码：

#include<iostream>#include<algorithm>using namespace std;int main(){int t,k,i,j,sum,n;int s[10001],num[10001];double max=-999999,temp,result;double a,b,aa,bb;double e=0.000001;int mark=0;cin>>t;for(k=1;k<=t;k++){cin>>n;sum=0;temp=0;max=0;result=0;for(i=0;i<n;i++){cin>>num[i];temp+=num[i]*num[i];}sort(num,num+n);for(i=0;i<n;i++){sum+=num[i];s[i]=sum;}for(i=0;i<n-1;i++){a=1.0*s[i]/(i+1);b=1.0*(sum-s[i])/(n-i-1);//cout<<"*"<<a<<" "<<b<<endl;result=temp+1.0*((i+1)*b*b+(n-i-1)*a*a-2*a*b*n);if(result>max){//cout<<i<<" "<<temp<<endl;max=result;mark=i;aa=a;bb=b;//cout<<aa<<bb<<endl;}}        printf("Case #%d: %.2lf\n",k,max);}}

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