HDU_1421DP

经典DP问题，先将劳累度排序，相邻相减的平方才是最小的，然后分析：dp[i][j]为前i物品选择了j对最小的疲劳值，对于dp[i][j] 当i = j*2时，则dp[i-2][j-1] + a, 当i>j*2时，dp[i][j] = min(dp[i-1][j], dp[i-2][j-1]+a)

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int N = 2000 + 10;

int w[N], dp[N][N/2];

int main()
{
    int n, k;

    while(~scanf("%d%d", &n, &k))
    {
        for(int i=1; i<=n; ++i)
        {
            scanf("%d", &w[i]);
        }

        sort(w+1, w+n+1);
        memset(dp, 0, sizeof(dp));
        for(int j=1; j<=k; ++j)
        {
            for(int i=j*2; i<=n; ++i)
            {
                int a = w[i]-w[i-1]; a *= a;
                if(j*2 == i)
                    dp[i][j] = dp[i-2][j-1] + a;

                else
                    dp[i][j] = min(dp[i-1][j], dp[i-2][j-1] + a);
            }
        }

        printf("%d\n", dp[n][k]);
    }

    return 0;
}