571B. Minimization(Codeforces Round #317)

B. Minimization

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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

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.

解題思路：

      將數組分成k組，分別是i,i+k,i+2*k,i+3*k...(1<=i<=k),有x=n%k個組元素個數是n/k+1個，問題就轉化爲k組內

相鄰元素差值的和的最小值，這時就需要對數組進行排序，只有每組內的元素都是有序的，每組內的相鄰元素的

差值纔會最小，接着就是在k組內分x組長度爲n/k+1，這時就需要dp，dp[i][j]，i是分了i組，j組長度是n/k+1;dp方

程爲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

從分第i組是,第i組的第1個元素與第i-1組的最後一個元素的差值。

代碼：

#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;
}