南夫拉斯（nfls）day？+1

A，B，C，D水，跳

E枚举中间点，一边桶存，一边遍历就好了，在写个二维前缀和就过了

F比较有意思，发现n较小，还明显可以贪心，贪心条件就只剩下最后的票数，于是枚举最后票数，贪心

代码

#include <bits/stdc++.h>
using namespace std;
long long n, ls, ans = 114514, piao, cnt[2005], res, cc[2005], lp;
struct sd {
    long long a, b;
} stu[2005];
int cmp(sd A, sd B) {
    if (B.b != A.b)
        return A.b < B.b;
    else
        return cnt[A.a] > cnt[B.a];
}
int main() {
    cin >> n;
    for (int i = 2; i <= n; i++) {
        cin >> stu[i].a;
        cnt[stu[i].a]++;
        if (stu[i].a == 1)
            piao++;
    }
    for (int i = 2; i <= n; i++) {
        cin >> stu[i].b;
    }
    sort(stu + 2, stu + n + 1, cmp);
    for (int i = piao; i <= n; i++) {
        res = 0;
        lp = piao;
        for (int j = 2; j <= n; j++) {
            if (cnt[stu[j].a] >= i && stu[j].a != 1) {
                cnt[stu[j].a]--;
                res += stu[j].b;
                cc[j] = 1;
                lp++;
            }
        }
        for (int j = 2; j <= n; j++) {
            if (lp >= i)
                break;
            if (cc[j] == 0 && stu[j].a != 1) {
                cnt[stu[j].a]--;
                res += stu[j].b;
                cc[j] = 1;
                lp++;
            }
        }
        ans = min(ans, res);
        for (int j = 2; j <= n; j++) {
            if (cc[j] == 1) {
                cnt[stu[j].a]++;
                cc[j] = 0;
            }
        }
    }
    cout << ans;
    return 0;
}