先对a[]按小到大排序.设:dp[i][j]表示前 i 个物品中搬 j
对的最少疲劳度.
1、当i==2*j时,dp[i][j]=dp[i-2][j-1]+(a[i]-a[i-1])^2
因为,当 (a1-a2)^2+(a3-a4)^2 <= (a1-a4)^2+(a3-a2)^2 ( a1<=a2<=a3<=a4 ).
2、当i>2*j时,dp[i][j]=min( dp[i-1][j] , dp[i-2][j-1]+(a[i]-a[i-1])^2 )
1 #include<cstdio>
2 #include<cstring>
3 #include<algorithm>
4 using namespace std;
5 int dp[2001][2001];
6 int w[2001];
7 int min(int a, int b)
8 {
9 return a<b?a:b;
10 }
11 int main()
12 {
13 int n,k,i,j;
14 while(scanf("%d%d",&n,&k)==2)
15 {
16 for(i=1;i<=n;i++)
17 scanf("%d",&w[i]);
18 sort(w+1,w+n+1);
19 for(i=0;i<=n;i++)
20 for(j=0;j<=k;j++)
21 dp[i][j] = 0;
22 for(i=2;i<=n;i++)
23 for(j=1;j*2<=i;j++)
24 {
25 if(i==2*j)
26 dp[i][j] = dp[i-2][j-1] + (w[i]-w[i-1])*(w[i]-w[i-1]);
27 else
28 dp[i][j] = min(dp[i-1][j],dp[i-2][j-1]+(w[i]-w[i-1])*(w[i]-w[i-1]));
29 }
30 printf("%d\n",dp[n][k]);
31 }
32 return 0;
33 }
参考：http://blog.csdn.net/pygzx/article/details/7381833