HDU1421

//dp 给定n个数字和k,求找出k对数字使得所有对数字差的平方和最小
//先给其从小到大进行排序,可知选取规则必须是选取相邻的两个数
//状态 dp[i][j]表示从前i个数选j组差的平方和最小
//当j*2==i时即i个数必须选完dp[i][j]=dp[i-2][j-1]+(a[i-1]-a[i])*(a[i-1]-a[i])
//否则dp[i][j]=min(dp[i-1][j],dp[i-2][j-1]+(a[i-1]-a[i])*(a[i-1]-a[i]))

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#include <iostream> #include <algorithm> #include <string.h> using namespace std; int a[2005]; long long dp[2005][1005]; int main() {     int n,k;     while(cin>>n>>k)     {         for(int i=1;i<=n;i++)         {             cin>>a[i];         }         sort(a+1,a+n+1);         for(int i=1;i<=2002;i++)             for(int j=1;j<=1002;j++)                 dp[i][j]=0;         for(int i=2;i<=n;i++)         {             for(int j=1;j<=k&&2*j<=i;j++)             {                 if(j*2==i)                     dp[i][j]=dp[i-2][j-1]+(a[i]-a[i-1])*(a[i]-a[i-1]);                 else                     dp[i][j]=min(dp[i-1][j],dp[i-2][j-1]+(a[i]-a[i-1])*(a[i]-a[i-1]));             }         }         cout<<dp[n][k]<<endl;     }     return 0; }