571B. Minimization(Codeforces Round #317)

You've got array A, consisting of n integers and a positive integer k. Array A is indexed by integers from 1 to n.

You need to permute the array elements so that value

$\text{[math]}$ became minimal possible. In particular, it is allowed not to change order of elements at all.

Input

The first line contains two integers n, k (2 ≤ n ≤ 3·105, 1 ≤ k ≤ min(5000, n - 1)).

The second line contains n integers A[1], A[2], ..., A[n] ( - 109 ≤ A[i] ≤ 109), separate by spaces — elements of the array A.

Output

Print the minimum possible value of the sum described in the statement.

Sample test(s)

input
3 2
1 2 4

output
1

input
5 2
3 -5 3 -5 3

output
0

input
6 3
4 3 4 3 2 5

output
3

Note

In the first test one of the optimal permutations is 1 4 2.

In the second test the initial order is optimal.

In the third test one of the optimal permutations is 2 3 4 4 3 5.

解题思路：

将数组分成k组，各自是i,i+k,i+2*k,i+3*k...(1<=i<=k),有x=n%k个组元素个数是n/k+1个，问题就转化为k组内

相邻元素差值的和的最小值，这时就须要对数组进行排序。仅仅有每组内的元素都是有序的，每组内的相邻元素的

差值才会最小，接着就是在k组内分x组长度为n/k+1，这时就须要dp，dp[i][j]。i是分了i组。j组长度是n/k+1;dp方

程为dp[i][j]=max(dp[i-1][j]+dp[i-1][j-1])+(a[i*n/k+j+1]-a[i*n/k+j])，ans=a[n]-a[1]-dp[k][x],a[i*n/k+j+1]-a[i*n/k+j]是要

从分第i组是,第i组的第1个元素与第i-1组的最后一个元素的差值。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn=500000+100;
int a[maxn];
int s[maxn];
int bbs(int x)
{
    if(x<0)
    return -x;
    return x;
}
int dp[5500][5500];
int main()
{
    int n,k;
    scanf("%d%d",&n,&k);
    for(int i=1;i<=n;i++)
    scanf("%d",&a[i]);
    sort(a+1,a+n+1);
    memset(dp,-1,sizeof(dp));
    int sum=a[n]-a[1];
    int q=n/k;
    int x=n%k;
    dp[0][0]=0;
    for(int i=1;i<=k;i++)
    {
        for(int j=0;j<=x;j++)
        {
            int df;
            if(j==0)
            df=dp[i-1][j];
            else
            df=max(dp[i-1][j],dp[i-1][j-1]);
            if(df<0)
            continue;
            if(i==k&&j==x)
            dp[i][j]=df;
            else
            dp[i][j]=df+a[i*q+j+1]-a[i*q+j];
        }
    }
    int ans=sum-dp[k][x];
    cout<<ans<<endl;
    return 0;
}