hdu1421搬寝室DP（ACM题目，动态规划）

分析：dp[i][j]表示前i个元素中 j 对数的最小差平方和

      dp[i][j] = min(dp[i-1][j], dp[i-2][j-1] + (weight[i] - weight[i-1])*(weight[i] - weight[i-1]));

code:

#include
#include

#define min(a,b) ((a) > (b) ? (b) : (a))
#define MaxAllow 2147483648
const int num = 2001;

int cmp(const void* a, const void *b)
{
 return *(int*)a - *(int*)b;
}

int dp[num][1001];  //数组要在函数外定义，不然太大不行，无语，又卡了n久

int main()
{
 int n, k;
 int i, j;
 int weight[num];
 while(scanf("%d%d", &n, &k) != EOF)
 {
  for(i = 1; i <= n; i++)
      scanf("%d", &weight[i]);

  qsort(weight+1, n, sizeof(weight[0]), cmp);
  for(i = 0; i <= n; i++)
      for(j = 1; j <= k; j++)
    dp[i][j] = MaxAllow;
        dp[0][0] = 0;
  
  for(i = 2; i <= n; i++)
      for(j = 1; j*2 <= i; j++)
      {
     dp[i][j] = min(dp[i-1][j],dp[i-2][j-1]+(weight[i]-weight[i-1])*(weight[i] - weight[i-1]));
   }
  printf("%d\n", dp[n][k]);
 }
 return 0;
}