第五届华中区程序设计邀请赛暨武汉大学第十四届校赛 网络预选赛 A

Problem 1603 - Minimum Sum

Time Limit: 2000MS   Memory Limit: 65536KB  
Total Submit: 564  Accepted: 157  Special Judge: No

Description

There are n numbers A[1] , A[2] .... A[n], you can select m numbers of it A[B[1]] , A[B[2]] ... A[B[m]]  ( 1 <= B[1] < B[2] .... B[m] <= n ) such that Sum as small as possible.

Sum is sum of abs( A[B[i]]-A[B[j]] ) when 1 <= i < j <= m.

Input

There are multiple test cases.
First line of each case contains two integers n and m.( 1 <= m <= n <= 100000 )
Next line contains n integers A[1] , A[2] .... A[n].( 0 <= A[i] <= 100000 )
It's guaranteed that the sum of n is not larger than 1000000.

Output

For each test case, output minimum Sum in a line.

Sample Input

4 2
5 1 7 10
5 3
1 8 6 3 10

Sample Output

2
8

 

题意： 长度为n的数组A 任意取 取出的数量为m 然后求   所有 abs( A[B[i]]-A[B[j]]  i<j ) 的和

题解： 比赛的时候 有一个思路  题目时间限制是2000ms 但woj 测评 2000+ms   过了

         先sort排序 取连续的长度为m的序列re[]   求其sum值 

     预处理求 re[1]-re[2] =ans[2]  并记录前缀和

                 re[1]-re[3] =ans[3]

                ...

                 re[1]-re[m]=ans[m]

    re[2]-re[3]=(re[1]-re[3])-(re[1]-re[2]) //

    re[2]-re[4]=(re[1]-re[4])-(re[1]-re[2]) //将要求的量 转化为已知量

 

#include<iostream>
 #include<cstdio>
 #include<cmath>
 #include<algorithm>
 #include<cstring>
 using namespace std;
int n,m;
struct node
 {
     long long  w;
     int pos;
 }N[100005];
long long ans[100005];
long long  re[100005];
long long  summ[100005];
long long qq;
long long exm;
long long gg;
int l=0,r=0;
bool cmp1(struct node aa,struct node bb)
 {
     if(aa.w<bb.w)
     return true;
     return false;
 }
long long fun( int ll,int rr)
 { 
       l=ll;
       r=rr;
       int jishu=1;
       for(int i=l;i<=r;i++)
           re[jishu++]=N[i].w;
           exm=0;
           summ[1]=0;
           ans[1]=0;
         for(int i=2;i<jishu;i++)
           { 
             ans[i]=abs(re[1]-re[i]);
             exm+=ans[i];
             summ[i]=exm;
           }
           qq=0;
         for(int i=1;i<jishu;i++)
         {
             qq=qq+summ[jishu-1]-summ[i];
             qq=qq-ans[i]*(jishu-i-1);
         }
     return qq;
 }
int main()
 {
     while(scanf("%d %d",&n,&m)!=EOF)
     {
         memset(N,0,sizeof(N));
         for(int i=1;i<=n;i++)
         {
           scanf("%lld",&N[i].w);
           N[i].pos=i;
         }
         sort(N+1,N+1+n,cmp1);
         gg=fun(1,m); 
         for(int i=2;i<=n-m+1;i++)
         {
             gg=min(gg,fun(i,i+m-1));  
         }
           printf("%lld
",gg);    
     }
     
 }