HDU-1421：搬寝室

/*状态方程：dp[i][j] =  min{ dp[i-1][j], dp[i-2][j-1]+(w[i]-w[i-1],2)^2 },0<j*2<=i其中dp[i][j]表示前i个物品拿j对在第i个物品的有两种情况，一：第i-1个物品和前面的物品恰好形成j对 二：第i-1个物品没有形成j对，由i-1个物品和第i个物品形成j对*/#include<cstdio>#include<algorithm>#include<iostream>#include<cmath>using namespace std;#define N 2000+5int dp[N][N];int w[N];int min(int a,int b){return a>b?b:a;}int main(){int n,k,i,j;while(scanf("%d%d",&n,&k)!=EOF){memset(dp,0,sizeof(dp));for(i=1;i<=n;i++)scanf("%d",&w[i]);sort(w+1,w+n+1);for(i=2;i<=n;i++){for(j=1;j*2<=i;j++){  if(j*2==i) dp[i][j] = dp[i-2][j-1]+pow(w[i]-w[i-1],2); else dp[i][j] = min(dp[i-1][j],dp[i-2][j-1]+pow(w[i]-w[i-1],2));}}printf("%d\n",dp[n][k]);}return 0;}