Educational Codeforces Round 175 (Rated for Div. 2) D. Tree Jumps

D. Tree Jumps

题目链接👈

题目描述🥰

题目思路😀

由于这个题目思路比较简单，所以就简单讲下，详细可以看代码理解

今天的题实际上就是一个典型的bfs序问题

根据题目上的定义，我们把1（即是根节点）的层数为1，每一层的节点除了不能接在父母的节点上面，都可以接在上一层的每一个节点的序列的后面，所以这个就有点bfs的思想，我们就从第一层开始推即可，一直推到最大层数即可输出答案。

AC代码🧠

// Problem: D. Tree Jumps
// Contest: Codeforces - Educational Codeforces Round 175 (Rated for Div. 2)
// URL: https://codeforces.com/contest/2070/problem/D
// Memory Limit: 512 MB
// Time Limit: 2000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

#include <bits/stdc++.h>

#define dev1(a) cout << #a << '=' << a << endl;
#define dev2(a, b) cout << #a << " = " << a << "  " << #b << " = " << b << endl;
#define dev3(a, b, c) cout << #a << " = " << a << "  " << #b << " = " << b << "  " << #c << " = " << c << endl;
#define dev4(a, b, c, d) cout << #a << " = " << a << "  " << #b << " = " << b << "  " << #c << " = " << c << "  " << #d << " = " << d << endl;
#define dev5(a, b, c, d, e) cout << #a << " = " << a << "  " << #b << " = " << b << "  " << #c << " = " << c << "  " << #d << " = " << d << "  " << #e << " = " << e << endl;
#define vec(a)                         \
    for (int i = 0; i < a.size(); i++) \
        cout << a[i] << ' ';           \
    cout << endl;
#define darr(a, _i, _n)               \
    cout << #a << ':';                \
    for (int ij = _i; ij <= _n; ij++) \
        cout << a[ij] << ' ';         \
    cout << endl;           
#define cin(a,n)           \
     for(int i=0;i<n;i++) \
      cin>>a[i];          
#define endl "\n"
#define pow pim
int pim(int a,int k)
{
    int res=1;
    if(a==0)return 0;
    while(k) 
    {
        if(k&1)res=(int)res*a;
        k>>=1;
        a=(int)a*a;
    }
    return res;
}
#define fi first
#define se second
#define caseT \
    int T;    \
    cin >> T; \
    while (T--)
#define int long long
// #define int __int128

using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> PII;
typedef pair<LL, LL> PLL;

const double PI = acos(-1.0);
const int INF = 0x3f3f3f3f;
const int MOD = 998244353;

const int N = 3e5+10;
int gcd(int a, int b)
{
    return b ? gcd(b, a % b) : a;
}
 
int lcm(int a, int b)
{
    return a * b / gcd(a, b);
}
inline int read()
{
    char c=getchar();int x=0,f=1;
    while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
    while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}
    return x*f;
}
int dep[N];
void solve()
{
	int n;
	cin>>n;
	int mx=0;
	dep[1]=1;
	map<int,int>hash;
	for(int i=2;i<=n;i++){
		int x;
		cin>>x;
		dep[i]=dep[x]+1;
		hash[dep[i]]++;
		mx=max(mx,dep[i]);
	}
	int ans=1;
	int pre=1;
	for(int i=2;i<=mx;i++)
	{
		ans+=(pre*hash[i]);
		pre*=(hash[i]-1);
		pre%=MOD;
		ans%=MOD;
	}
	cout<<ans<<endl;
}

signed main()
{

    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
	caseT
    solve();
    return 0;
}
/*

*/