hdu3480 Division

Problem Description

Little D is really interested in the theorem of sets recently. There’s a problem that confused him a long time.  
Let T be a set of integers. Let the MIN be the minimum integer in T and MAX be the maximum, then the cost of set T if defined as (MAX – MIN)^2. Now given an integer set S, we want to find out M subsets S1, S2, …, SM of S, such that



and the total cost of each subset is minimal.

 

Input

The input contains multiple test cases.
In the first line of the input there’s an integer T which is the number of test cases. Then the description of T test cases will be given. 
For any test case, the first line contains two integers N (≤ 10,000) and M (≤ 5,000). N is the number of elements in S (may be duplicated). M is the number of subsets that we want to get. In the next line, there will be N integers giving set S.

 

Output

For each test case, output one line containing exactly one integer, the minimal total cost. Take a look at the sample output for format.

 

Sample Input

2 3 2 1 2 4 4 2 4 7 10 1

 

Sample Output

Case 1: 1 Case 2: 18

题意：给你n个数字，要分成m个子集合，子集合之间可以共用相同的元素，子集合的代价为该集合最大的数和最小的数的差的平方，问你最小的子集合的总代价是多少。

思路：因为集合里的数的顺序并不是按顺序的，所以可以先排序，然后用dp[i][j]表示前i个数分成j个集合所得到的最小代价，和邮局那题有点相似，要用四边形优化，不然会超时，看别人的写法发现斜率优化更快，学完后再来写吧。

#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<string>
#include<algorithm>
using namespace std;
#define ll long long
#define inf 999999999
int a[10060],dp[10060][5060],s[10060][5060];
int main()
{
    int n,m,i,j,T,len,k,num1=0;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&m);
        for(i=1;i<=n;i++){
            scanf("%d",&a[i]);
        }
        sort(a+1,a+1+n);
        for(i=1;i<=n;i++){
            dp[i][1]=(a[i]-a[1])*(a[i]-a[1]);
            s[i][1]=1;
        }
        for(j=2;j<=m;j++){
            dp[j][j]=0;
            s[n+1][j]=n;
            for(i=n;i>j;i--){
                dp[i][j]=inf;
                for(k=s[i][j-1];k<=s[i+1][j];k++){
                    if(dp[i][j]>dp[k][j-1]+(a[i]-a[k+1])*(a[i]-a[k+1])){
                        dp[i][j]=dp[k][j-1]+(a[i]-a[k+1])*(a[i]-a[k+1]);
                        s[i][j]=k;
                    }
                }
            }
        }
        num1++;
        printf("Case %d: %d
",num1,dp[n][m]);
    }
    return 0;
}