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
![](http://acm.hdu.edu.cn/data/images/C295-1003-1.jpg)
and the total cost of each subset is minimal.
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
![](http://acm.hdu.edu.cn/data/images/C295-1003-1.jpg)
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.
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;
}