bnu 51640 Training Plan DP

https://www.bnuoj.com/bnuoj/problem_show.php?pid=51640

dp[i][j]表示前j个数，分成了i组，最小需要多少精力。

那么，求解订票dp[i][j]的时候，要么，第i组不做题，要么，第i组做1题、2题、3题....j题

先把数组排好序。然后暴力

dp[i][j] = dp[i - 1][j] //第i组不做题。

然后枚举一个k，表示[K.....j]是第i组做的题。那么就是dp[i - 1][k - 1] + (a[j] - a[k])^2

复杂度是O（n^3）

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <assert.h>
#define IOS ios::sync_with_stdio(false)
using namespace std;
#define inf (0x3f3f3f3f)
typedef long long int LL;

#include <iostream>
#include <sstream>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <bitset>
const int maxn =  + ;
int a[maxn];
LL dp[maxn][maxn];
void work() {
    int n, m;
    scanf("%d%d", &n, &m);
    for (int i = ; i <= n; ++i) {
        scanf("%d", &a[i]);
    }
    sort(a + , a +  + n);
    for (int i = ; i <= n; ++i) dp[][i] = 1e18;
    dp[][] = ;
    for (int i = ; i <= m; ++i) {
        for (int j = ; j <= n; ++j) {
            dp[i][j] = dp[i - ][j];
            for (int k = ; k <= j; ++k) {
                dp[i][j] = min(dp[i][j], dp[i - ][k - ] + 1LL * (a[j] - a[k]) * (a[j] - a[k]));
            }
        }
    }
    printf("%lld\n", dp[m][n]);
}

int main() {
#ifdef local
    freopen("data.txt", "r", stdin);
//    freopen("data.txt", "w", stdout);
#endif
    int t;
    scanf("%d", &t);
    while (t--) work();
    return ;
}