HDU 1421 Dynamic Programming

Title:

There are n items, from which 2*k items are selected, so that the sum of the squares of the difference between the weights of the two items in each group of these k groups is the smallest

answer:

Sort n items first, and try to select an item adjacent to it to pair it. 

dp[i][j] means to select i pair from the first j items

Then for the jth item, there are two choices: choose or not, that is, dp[i][j-1] or dp[i-1][j-2]+(a[j]-a[j- 1])^2

#include<cstdio>
#include<cstring>
#include<string>
#include<iostream>
#include<sstream>
#include<algorithm>
#include<utility>
#include<vector>
#include<set>
#include<map>
#include<queue>
#include<cmath>
#include<iterator>
#include<stack>
using namespace std;
const int INF=0x3f3f3f3f;
const double eps=1e-7;
const int mod=1000007;
const int maxn=2005;
int dp[maxn][maxn];
int a[maxn];
intmain()
{
        int n,k;
      while(cin>>n>>k)
      {
              for(int i=1;i<=n;i++)
                cin>>a[i];
              sort(a+1,a+1+n);
              memset(dp,INF,sizeof dp);//Indicates that it cannot be reached
              for(int i=0;i<=n;i++)
                dp[0][i]=0;//No need to select any items
              for(int i=1;i<=k;i++)
              {
                for(int j=2;j<=n;j++)
                dp[i][j]=min(dp[i][j-1],dp[i-1][j-2]+(a[j]-a[j-1])*(a[j]-a[j-1]));
              }
              cout<<dp[k][n]<<endl;
      }
}