NOJ1581

题意：给出n只筷子，选出k+3双筷子，使得所有筷子的平方差的和最小

思路：我们可以从小到达排序，这样的话，每一只筷子与左右两只筷子的平方差会比与其他只筷子的平方差小。

所以d[i][j]表示前i只筷子组合成j对的最小平方差的和，对于第i只筷子就存在两种情况，取或者不取。取的话arr[i]就要与arr[i - 1]配对，d[i][j] = min(d[i - 1][j], d[i - 2][j - 1]  + (arr[i] - arr[i - 1]) ^ 2);不取的话，d[i][j] = d[i- 1][j]；

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

using namespace std;

const int MAXN = 105;
const int INF = 0x3f3f3f3f;

int arr[MAXN], d[MAXN][MAXN];
int n, k;

int main() {
    scanf("%d %d", &n, &k);
        k += 3;
    for (int i = 1; i <= n; i++) 
        scanf("%d", &arr[i]);
    sort(arr + 1, arr + 1 + n);

    for (int i = 0; i <= n; i++) {
        d[i][0] = 0;
        for (int j = 1; j <= k; j++)
            d[i][j] = INF;
    }

    for (int i = 2; i <= n; i++)
        for (int j = 1; j <= k; j++)
            d[i][j] = min(d[i - 1][j], d[i - 2][j - 1] + (arr[i] - arr[i - 1]) * (arr[i] - arr[i - 1]));

    if (n >= k * 2)
        printf("%d\n", d[n][k]);
    else
        printf("-1\n"); 

    return 0;
}