题目链接

题意：



题解：


















AC代码

/*
    Author : zzugzx
    Lang : C++
    Blog : blog.csdn.net/qq_43756519
*/
#include<bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(), (x).end()
#define endl '\n'
#define SZ(x) (int)x.size()
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
const int mod = 19260817;
//const int mod = 998244353;
const double eps = 1e-10;
const double pi = acos(-1.0);
const int maxn = 1e6+10;
const ll inf = 0x3f3f3f3f;
const int dir[][2]={{0, 1}, {1, 0}, {0, -1}, {-1, 0}, {1, 1}, {1, -1}, {-1, 1}, {-1, -1}};

ll fa[maxn], d[maxn], son[maxn], father[maxn], cnt[maxn];
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);cout.tie(0);
//  freopen("in.txt", "r", stdin);
//  freopen("out.txt", "w", stdout);
    int n, m;
    cin >> n >> m;
    for (int i = 2; i <= n; i++){
        int x;
        cin >> x;
        fa[i] = x;
        d[x]++; d[i]++;
    }
    ll ans = 0;
    for (int i = 1, x; i <= m; i++) {
        cin >> x;
        ++cnt[x];
        son[x] = (son[x] + d[x]) % mod;
        son[fa[x]] = (son[fa[x]] + 2) % mod;
        son[fa[fa[x]]] = (son[fa[fa[x]]] + 1) % mod;
        father[x] = (father[x] + 1) % mod;
        father[fa[x]] = (father[fa[x]] + 1) % mod;
        ans = (ans + i * (son[x] + father[fa[x]] + cnt[fa[x]] + cnt[fa[fa[x]]])) % mod;
    }
    cout << ans;
    return 0;
}