結果

問題 No.3343 Distance Sum of Large Tree
コンテスト
ユーザー GOTKAKO
提出日時 2025-11-13 22:43:31
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
WA  
実行時間 -
コード長 3,880 bytes
コンパイル時間 2,163 ms
コンパイル使用メモリ 211,720 KB
実行使用メモリ 8,876 KB
最終ジャッジ日時 2025-11-13 22:43:35
合計ジャッジ時間 4,353 ms
ジャッジサーバーID
（参考情報） 		judge5 / judge1
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 2
other AC * 4 WA * 26
権限があれば一括ダウンロードができます

ソースコード

diff # 

#include <bits/stdc++.h>
using namespace std;

//入力が必ず-mod<=a<modの時.
template<const int mod> //mod<2^30.
struct modint{ //mod変更が不可能.
    public:
    long long v;
    static void setmod(int m){} //飾り.
    static constexpr long long getmod(){return mod;}
    modint():v(0){}
    template<typename T>
    modint(T a):v(a){if(v < 0) v += mod;}
    long long val()const{return v;}
 
    modint &operator=(const modint &b) = default;
    modint &operator+()const{return (*this);}
    modint operator-()const{return modint(0)-(*this);}
    modint operator+(const modint b)const{return modint(v)+=b;}
    modint operator-(const modint b)const{return modint(v)-=b;}
    modint operator*(const modint b)const{return modint(v)*=b;}
    modint operator/(const modint b)const{return modint(v)/=b;}
    modint &operator+=(const modint b){
        v += b.v; if(v >= mod) v -= mod;
        return *this;
    }
    modint &operator-=(const modint b){
        v -= b.v; if(v < 0) v += mod; 
        return *this;
    }   
    modint &operator*=(const modint b){v = v*b.v%mod; return *this;}
    modint &operator/=(modint b){ //b!=0 mod素数が必須.
        assert(b.v != 0);
        (*this) *= b.pow(mod-2);
        return *this;
    }
    modint pow(long long n)const{
        modint ret = 1,p = v;
        if(n < 0) p = p.inv(),n = -n;
        while(n){
            if(n&1) ret *= p;
            p *= p; n >>= 1;
        }
        return ret;
    }
    modint inv()const{return pow(mod-2);} //素数mod必須.
 
    modint &operator++(){*this += 1; return *this;}
    modint &operator--(){*this -= 1; return *this;}
    modint operator++(int){modint ret = *this; *this += 1; return ret;}
    modint operator--(int){modint ret = *this; *this -= 1; return ret;}
    friend bool operator==(const modint a,const modint b){return a.v==b.v;}
    friend bool operator!=(const modint a,const modint b){return a.v!=b.v;}
    friend bool operator<(const modint a,const modint b){return a.v<b.v;}
    friend bool operator<=(const modint a,const modint b){return a.v<=b.v;}
    friend bool operator>=(const modint a,const modint b){return a.v>=b.v;}
    friend bool operator>(const modint a,const modint b){return a.v>b.v;}
    friend ostream &operator<<(ostream &os,const modint a){return os<<a.v;}
    friend istream &operator>>(istream &is,modint &a){ //入力はmodをとってくれる.
        long long x; is >> x; x %= mod;
        a = modint(x); return is;
    }
};
using mint = modint<998244353>; const long long mod = 998244353;

int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);

    int N; cin >> N;
    vector<int> A(N),B(N-1),C(N-1),P(N-1);
    for(auto &a : A) cin >> a;
    for(auto &a : B) cin >> a,a--;
    for(auto &a : C) cin >> a,a--;
    for(auto &a : P) cin >> a,a--;

    vector<vector<tuple<int,int,int>>> Graph(N);
    for(int i=0; i<N-1; i++){
        int b = B.at(i),c = C.at(i),p = P.at(i);
        Graph.at(p).push_back({c,b,i+1});
    }
    for(auto &g : Graph) sort(g.begin(),g.end());
    mint answer = 0,div6 = mint(1)/6;
    auto dfs = [&](auto dfs,int pos,long long back) -> pair<mint,mint> {
        mint siz = 0,sum = 0;
        long long a = A.at(pos);
        answer += a*((a-1)*a/2%mod)%mod;
        answer -= div6*((a-1)*a%mod)*((2*a-1)%mod);
        auto f = [&](long long x) -> mint {return (x*(x+1)/2%mod + (a-x)*(a-x-1)/2%mod)%mod;};
        siz += a,sum += f(back);

        mint ksiz = 0,ksum = 0,ksum2 = 0;
        for(auto [b,c,to] : Graph.at(pos)){
            auto [k1,k2] = dfs(dfs,to,c);
            answer += k2*a+k1*f(b);
            siz += k1,sum += k2+k1*(abs(b-back)%mod);
            answer += ksiz*b-ksum;
            answer += ksiz*k2+ksum2*k1;
            ksiz += k1,ksum += k1*b,ksum2 += k2;
        }
        return {siz,sum+siz};
    };
    dfs(dfs,0,0);
    cout << answer*2 << "\n";
}
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