codeforces 571B Minimization(dp)

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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 21 2 4
output
1
input
5 23 -5 3 -5 3
output
0
input
6 34 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.

题目链接http://codeforces.com/contest/572/problem/D

计算|arr[1]-arr[k+1]|+|arr[2]-arr[k+2]|+......+|arr[n-k]-arr[n]|,不妨将前式转换为

+|arr[2]-arr[k+2]|+|arr[k+2]-arr[k+2+k]|+|arr[k+2+k]-arr[k+2+k+k]|+....(其中k+2+k+...+k<=数组元素的个数n)

+......

+|arr[k]-arr[k+k]|+|arr[k+k]-arr[k+k+k]|+|arr[k+k+k]-arr[k+k+k+k]|+....(其中k+k+k+...+k<=数组元素的个数n)

这样分成了以1,2....k开头的k组;每组的长度可能是n/k或者n/k+1;

长度为n/k+1的组数为n%k;长度为n/k的组数为k-n%k;

只有每组的元素都是有序的,每组的相邻元素的差的和才会最小;

若每组元素都是有序的,则上面k组的结果就分别等价于:

|arr[1]-arr[k+1+k+.....+k]| ,|arr[2]-arr[k+2+k+...k]| , ....... ,|arr[k]-arr[k+k+k+...k]|;

这样每组的和只与这一组的第一个元素和最后一个元素有关; 类似于((a1-a2)+(a2-a3)+...+(a[n-1]-a[n]))=a1-a[n];

最终结果就是这k组的和;

所以现在的问题就是这k组中n%k个较长的组和k-n%k个较短的组怎么选;

比如第二组数据 -5,-5,3,3,3 ;其中长组的个数为1,短组的个数为1;段组的长度为2;长组的长度为3;

先对数据排序,如果选-5,-5,3为一组(长组),3,3为一组(短组);那么最后结果就是错的;

所以需要递推去分组;

我也是看了别人的程序后才彻底明白的,具体看代码;

#include<iostream>#include<cstdio>#include<cstring>#include<string>#include<algorithm>#include<map>#include<queue>#include<climits>#include<list>#include<stack>#include<cmath>#define ll long long#define MAX 310000#define mem(a)  memset(a,0,sizeof(a))#define mems(a)  memset(a,-1,sizeof(a))using namespace std;int arr[MAX];ll dp[5500][5500];int main(){    int n,k;    scanf("%d%d",&n,&k);    int i;    for(i=1;i<=n;++i)    {        scanf("%d",&arr[i]);    }    sort(arr+1,arr+n+1);    int j;    dp[0][0]=0;    int num_long,num_short;    num_long=n%k;//较长组的个数    num_short=k-num_long;//较短组的个数    int len=n/k;//短组的长度    for(i=0;i<=num_long;++i)    {        for(j=0;j<=num_short;++j)        {            if(i==0&&j==0)            {                continue;            }            int pos=i*(len+1)+j*len;//当前分了多少个,也是当前所分的组的最后一个元素在arr中的坐标            dp[i][j]=INT_MAX;            if(j>0)            {                dp[i][j]=min(dp[i][j],dp[i][j-1]+arr[pos]-arr[pos-len+1]);            }            if(i>0)            {                dp[i][j]=min(dp[i][j],dp[i-1][j]+arr[pos]-arr[pos-len]);            }        }    }    printf("%lld\n",dp[num_long][num_short]);}

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