题目描述
给你一棵树，最开始点权为0，每次将与一个点x树上距离<=1的所有点点权+1，之后询问这些点修改后的点权和.
输入描述:
第一行两个数n和m
第二行n-1个数，第i个数fa[i + 1]表示i + 1点的父亲编号，保证fa[i + 1]<i + 1
第三行m个数，每个数x依次表示这次操作的点是x
输出描述:
输出一个数，即这m次操作的答案的hash值
如果是第i次操作，这次操作结果为ans,则这个hash值加上
i * ans
输出hash值对19260817取模的结果

示例1
输入
6 3
1 1 2 3 3
1 2 3
输出
34
示例2
输入
6 10
1 1 2 3 3
1 4 6 5 2 3 3 3 3 3
输出
869

解题思路

#pragma GCC target("avx,sse2,sse3,sse4,popcnt")
#pragma GCC optimize("O2,O3,Ofast,inline,unroll-all-loops,-ffast-math")
#include <bits/stdc++.h>
using namespace std;
#define js ios::sync_with_stdio(false);cin.tie(0); cout.tie(0)
#define all(__vv__) (__vv__).begin(), (__vv__).end()
#define endl "\n"
#define pai pair<int, int>
#define mk(__x__,__y__) make_pair(__x__,__y__)
#define ms(__x__,__val__) memset(__x__, __val__, sizeof(__x__))
typedef long long ll; typedef unsigned long long ull; typedef long double ld;
const int MOD = 19260817;
const int INF = 0x3f3f3f3f;
inline ll read() { ll s = 0, w = 1; char ch = getchar(); for (; !isdigit(ch); ch = getchar()) if (ch == '-') w = -1; for (; isdigit(ch); ch = getchar())    s = (s << 1) + (s << 3) + (ch ^ 48); return s * w; }
inline void print(ll x) { if (!x) { putchar('0'); return; } char F[40]; ll tmp = x > 0 ? x : -x; if (x < 0)putchar('-');    int cnt = 0;    while (tmp > 0) { F[cnt++] = tmp % 10 + '0';        tmp /= 10; }    while (cnt > 0)putchar(F[--cnt]); }
inline ll gcd(ll x, ll y) { return y ? gcd(y, x % y) : x; }
ll qpow(ll a, ll b) { ll ans = 1;    while (b) { if (b & 1)    ans *= a;        b >>= 1;        a *= a; }    return ans; }    ll qpow(ll a, ll b, ll mod) { ll ans = 1; while (b) { if (b & 1)(ans *= a) %= mod; b >>= 1; (a *= a) %= mod; }return ans % mod; }
inline int lowbit(int x) { return x & (-x); }

const int N = 1e5 + 7;
ll son[N], father[N], fa[N];
int d[N], cnt[N];

int main() {
    int n = read(), m = read();
    for (int i = 2; i <= n; ++i) {
        int x = read();
        fa[i] = x;
        ++d[x], ++d[i];
    }
    ll ans = 0;
    for (int i = 1; i <= m; ++i) {
        int x = read();
        ++cnt[x];    // x节点的操作次数+1
        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) % MOD;
    }
    print(ans);
    return 0;
}