RAZLIKA

Description

Mirko's newest math homework assignment is a very difficult one! Given a sequence, V, of N integers,
remove exactly K of them from the sequence. Let M be the largest difference of any two remaining
numbers in the sequence, and m the smallest such difference. Select the K integers to be removed from
V in such a way that the sum M + m is the smallest possible. Mirko isn't very good at math, so he has
asked you to help him!

Input

The first line of input contains two positive integers, N (3 ≤ N ≤ 1 000 000) and K (1 ≤ K ≤ N - 2).
The second line of input contains N space-separated positive integers – the sequence V (-5 000 000 ≤
Vi ≤ 5 000 000).

Output

The first and only line of output must contain the smallest possible sum M + m

Sample Input

5 2
-3 -2 3 8 6

Sample Output
7

// 题目要求的是取出n-k个数，这些数的最大元素差+最小元素差 最小
// 这里明显是先排序，然后取其中的n-k个数，
// 我们可以枚举结尾 i ，这里先要预处理求出以 i 结尾的前n-k 个数之间的最小值、
#include <cstdio>
#include <algorithm>
#include<ctime>
#include <cstring>
#include<iostream>
#include<cmath>
#include<stack>
#define maxn 1000010
#define INF 30000000
#define LL long long
#define mod 100007
#define pi acos(-1.0)100099
using namespace std ;
#pragma comment(linker,"/STACK:102400000,102400000")
int a[maxn] ;
int min1[maxn][2] ,min2[maxn][2] ;
//min1[i][0] 保存的是以i结尾的前k个数两着之间的最小值,min2[i][0]是次小值
//min1[i][1] 是取得最小值时它的位置
int main()
{
   int  j , len , i , k ,jj ;
   int n , ans, m ;
//freopen("in.txt","r",stdin) ;
   while( scanf("%d%d" , &n ,&k ) != EOF  )
   {
        ans = INF ;
        for( i = 1 ; i <= n ; i++ )
         scanf("%d" , &a[i]) ;
         sort(a+1,a+1+n) ;
         min1[1][0] = INF ;min2[2][0] = INF ;
         min2[1][0] = INF ;min1[2][0] = a[2]-a[1] ;
         min1[2][1] = 1 ;min2[1][0] = 1 ;
         k = n-k ;
         for( i = 3 ; i <= n ;i++ )
         {
             j = min1[i-1][1];
             jj = min2[i-1][1] ;
             if( i - j >= k || i - jj >= k )
             {
                 if(i-j >= k)
                 {
                     if( a[i]-a[i-1] <= min2[i-1][0] )
                     {
                       min1[i][0] = a[i]-a[i-1] ;
                       min1[i][1] = i-1 ;
                       min2[i][0] = min2[i-1][0];
                       min2[i][1] = min2[i-1][1] ;
                     }
                     else
                     {
                         min1[i][0] = min2[i-1][0] ;
                         min1[i][1] = min2[i-1][1] ;
                         min2[i][0] = a[i]-a[i-1] ;
                         min2[i][1] = i-1 ;
                     }
                 }
                 else if(i-jj >= k)
                 {
                      if(a[i]-a[i-1] <= min1[i-1][0])
                      {
                        min2[i][0] = min1[i-1][0];
                        min2[i][1] = min1[i-1][1];
                        min1[i][0] = a[i]-a[i-1] ;
                        min1[i][1] = i-1 ;
                      }
                      else
                      {
                        min2[i][0] = a[i]-a[i-1] ;
                        min2[i][1] = i-1 ;
                        min1[i][0] = min1[i-1][0];
                        min1[i][1] = min1[i-1][1];
                      }
                 }
             }
             else
             {
                 if(a[i]-a[i-1] <= min1[i-1][0])
                 {
                     min2[i][0] = min1[i-1][0];
                     min2[i][1] = min1[i-1][1];
                     min1[i][0] = a[i]-a[i-1] ;
                     min1[i][1] = i-1 ;
                 }
                 else if( a[i]-a[i-1] <= min2[i-1][0] )
                 {
                     min2[i][0] = a[i]-a[i-1];
                     min2[i][1] = i-1;
                     min1[i][0] = min1[i-1][0] ;
                     min1[i][1] = min1[i-1][1] ;
                 }
                 else
                 {
                     min1[i][0] = min1[i-1][0] ;
                     min1[i][1] = min1[i-1][1] ;
                     min2[i][0] = min2[i-1][0] ;
                     min2[i][1] = min2[i-1][1] ;
                 }
             }
         }
         //枚举结尾
         for( i = k ; i <= n ;i++ )
         {
             jj = a[i]-a[i-k+1] ;
             if(jj+min1[i][0] < ans )
                ans = jj+min1[i][0] ;
         }
        printf("%d ",ans) ;
   }
   return 0 ;
}