Little D is really interested in the theorem of sets recently. There’s a problem that confused him a long time.  
Let T be a set of integers. Let the MIN be the minimum integer in T and MAX be the maximum, then the cost of set T if defined as (MAX – MIN)^2. Now given an integer set S, we want to find out M subsets S1, S2, …, SM of S, such that

and the total cost of each subset is minimal.

The input contains multiple test cases.
In the first line of the input there’s an integer T which is the number of test cases. Then the description of T test cases will be given.

For any test case, the first line contains two integers N (≤ 10,000) and M (≤ 5,000). N is the number of elements in S (may be duplicated). M is the number of subsets that we want to get. In the next line, there will be N integers giving set S.

For each test case, output one line containing exactly one integer, the minimal total cost. Take a look at the sample output for format.

2
3 2
1 2 4
4 2
4 7 10 1

Case 1: 1
Case 2: 18


Hint
The answer will fit into a 32-bit signed integer.
 

zhengfeng

#include <iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
#define MAX(a,b) ((a)>(b)?(a):(b))
#define MIN(a,b) ((a)<(b)?(a):(b))
#define positive(a) ((a)>0?(a):-(a))
using namespace std;

int a[10010],dp[10010][5010];//a存数值。dp[i][j]表示前i个元素分成j个子集的最小价值
int n,m,head,tail;
int q[10010];//队列
int main()
{
    int t,cas,i,j,k,x1,x2,x3,y1,y2,y3,p1,p2,p3;

    scanf("%d",&t);
    for(cas=1;cas<=t;cas++)
    {
        scanf("%d%d",&n,&m);
        for(i=1;i<=n;i++)
            scanf("%d",a+i);
        sort(a+1,a+n+1);
        for(i=1;i<=n;i++)
        {
            dp[i][1]=(a[i]-a[1])*(a[i]-a[1]);//一个子集即自己本身
            dp[i][i]=0;//全为单元集价值为0
        }
        for(j=2;j<=m;j++)
        {
            head=tail=0;
            q[tail++]=j-1;//k的范围为j-1<=k<i所以先入队两个元素
            q[tail++]=j;
            for(i=j+1;i<=n;i++)
            {
                while(tail-head>=2)
                {
                    p1=q[head];
                    p2=q[head+1];
                    y1=dp[p1][j-1]+a[p1+1]*a[p1+1];
                    x1=a[p1+1];
                    y2=dp[p2][j-1]+a[p2+1]*a[p2+1];
                    x2=a[p2+1];
                    if(y2-y1<=2*a[i]*(x2-x1))//(y2-y1)/(x2-x1)<=2*a[i]说明y2更优所以去掉y1
                        head++;
                    else
                        break;
                }
                k=q[head];
                dp[i][j]=dp[k][j-1]+(a[i]-a[k+1])*(a[i]-a[k+1]);
                while(tail-head>=2)//加入i去掉上凸点
                {
                    p1=q[tail-2];
                    p2=q[tail-1];
                    p3=i;
                    y3=dp[p3][j-1]+a[p3+1]*a[p3+1];
                    x3=a[p3+1];
                    y2=dp[p2][j-1]+a[p2+1]*a[p2+1];
                    x2=a[p2+1];
                    y1=dp[p1][j-1]+a[p1+1]*a[p1+1];
                    x1=a[p1+1];
                    if((y3-y2)*(x2-x1)<=(y2-y1)*(x3-x2))
                        tail--;
                    else
                        break;
                }
                q[tail++]=i;
            }
        }
        printf("Case %d: %d\n",cas,dp[n][m]);
    }
    return 0;
}