rt，思路是枚举所有从根节点出发，到达最底下的路径，然后用 s 计算，并排序，取最大的两个，再算出总贡献。

这是赛时想的，但是一直 TLE，赛后一分钟发现这个 sb 在 #define int long long 的情况下使用了 scanf("%d");

不知道会不会 WA 什么的，现在提交不了，代码：

#include <iostream>
#include <vector>
#include <algorithm>

using namespace std;
#define int long long

const int N = 200010;

int T;
int n, k;
int p[N], s[N];
vector<int> v[N];
vector<int> chain[N];
int cnt = 0;
int now[N];
int sum[N];

void dfs(int x, int len) {
    if (v[x].size() == 0) {
        cnt++; chain[cnt].push_back(1);
        for (int i = 1; i <= len; i++) chain[cnt].push_back(now[i]);
        return;
    }
    for (int i = 0; i < v[x].size(); i++) {
        int son = v[x][i];
        now[len + 1] = son;
        dfs(son, len + 1);
    }
}

signed main() {
    cin >> T;
    while (T--) {
        scanf("%lld%lld", &n, &k);
        for (int i = 1; i <= n; i++) v[i].clear(), chain[i].clear();
        for (int i = 2; i <= n; i++) {
            scanf("%lld", p + i);
            v[p[i]].push_back(i);
        }
        for (int i = 1; i <= n; i++) scanf("%lld", s + i);
        dfs(1, 0);
        /*
        for (int i = 1; i <= cnt; i++) {
            for (int j = 0; j < chain[i].size(); j++) cout << chain[i][j] << " ";
            cout << endl;
        }
        */
        for (int i = 1; i <= cnt; i++) {
            sum[i] = 0;
            for (int j = 0; j < chain[i].size(); j++) 
                sum[i] += s[chain[i][j]];
        }
        sort(sum + 1, sum + 1 + cnt);
        int a = sum[cnt], b = sum[cnt - 1];
        if (k % 2 == 0) cout << (a + b) * (k / 2) << "\n";
        else  cout << (k / 2 + 1) * a + (k / 2) * b << "\n";
    }
    return 0;
}