#812317 (C++23) No.2115 Making Forest Easy

• 問題: No.2115 Making Forest Easy
• ユーザー: 👑 rin204
• 提出日時: 2022-10-28 23:35:54
• 言語: C++23 (gcc 13.3.0 + boost 1.87.0)
• 結果: TLE
• 実行時間: -
• コード長: 3,531 bytes
• コンパイル時間: 5,405 ms
• コンパイル使用メモリ: 296,664 KB
• 実行使用メモリ: 6,944 KB
• 最終ジャッジ日時: 2024-07-06 02:53:38
• 合計ジャッジ時間: 34,478 ms
• ジャッジサーバーID （参考情報）: judge1 / judge5

ファイルパターン	結果
sample	AC * 2
other	AC * 49 TLE * 1

ソースコード

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#pragma GCC target("avx2")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
#include<bits/stdc++.h>
#include<atcoder/modint>
using namespace std;
using namespace atcoder;
using mint = modint998244353;
void solve(){
    int n;
    cin >> n;
    vector<pair<long long, int>> A(n);
    long long a;
    for(int i = 0; i < n; i++){
        cin >> a;
        A[i] = {a, i};
    }
    vector<vector<int>> edges(n, vector<int>());
    int u, v;
    for(int i = 0; i < n - 1; i++){
        cin >> u >> v;
        u--; v--;
        edges[u].push_back(v);
        edges[v].push_back(u);
    }
    sort(A.begin(), A.end(), greater<pair<long long, int>>());
    vector<bool> used(n, false);
    vector<mint> pow2(n + 1, 1);
    for(int i = 1; i <= n; i++) pow2[i] = pow2[i - 1] * 2;
    
    mint ans = 0;
    for(auto tmp:A){
        a = tmp.first;
        int i = tmp.second;
        vector<int> dist(n, -1);
        vector<int> size(n, 0);
        dist[i] = 0;
        stack<int> st;
        st.push(~i);
        st.push(i);
        while(!st.empty()){
            int pos = st.top();
            st.pop();
            if(pos >= 0){
                for(auto npos: edges[pos]){
                    if(dist[npos] == -1){
                        dist[npos] = dist[pos] + 1;
                        st.push(~npos);
                        st.push(npos);
                    }
                }
            }
            else{
                pos = ~pos;
                size[pos]++;
                for(auto npos: edges[pos]) size[pos] += size[npos];
            }
        }
        vector<mint> dp(n, 0);
        auto dfs=[&](auto self, int pos, int bpos) -> void {
            dp[pos] = 1;
            for(auto npos: edges[pos]){
                if(npos == bpos) continue;
                if(used[npos]){
                    dp[pos] *= pow2[size[npos] - 1];
                }
                else{
                    self(self, npos, pos);
                    dp[pos] *= dp[npos] + pow2[size[npos] - 1];
                }
            }
        };
        dfs(dfs, i, -1);
        
        mint tot = 0;
        
        auto dfs2=[&](auto self, int pos, int bpos) -> void {
            vector<mint> L(1, 1);
            for(auto npos: edges[pos]){
                if(used[npos]) L.push_back(L.back() * pow2[size[npos] - 1]);
                else if(npos == bpos) L.push_back(L.back() * dp[npos]);
                else L.push_back(L.back() * (pow2[size[npos] - 1] + dp[npos]));
            }
            tot += L.back();
            vector<mint> R(1, 1);
            
            for(int ii = edges[pos].size() - 1; ii >= 0; ii--){
                int npos = edges[pos][ii];
                if(used[npos]) R.push_back(R.back() * pow2[size[npos] - 1]);
                else if(npos == bpos) R.push_back(R.back() * dp[npos]);
                else R.push_back(R.back() * (pow2[size[npos] - 1] + dp[npos]));
            }
            
            reverse(R.begin(), R.end());
            for(int ii = 0; ii < edges[pos].size(); ii++){
                int npos = edges[pos][ii];
                if((npos == bpos) || used[npos]) continue;
                dp[pos] = L[ii] * R[ii + 1];
                size[pos] = n - size[npos];
                self(self, npos, pos);
            }
        };
        dfs2(dfs2, i, -1);
        ans += tot * a;
        used[i] = true;
    }
    cout << ans.val() << "\n";
}
int main(){
    cin.tie(0)->sync_with_stdio(0);
    int t;
    t = 1;
    // cin >> t;
    while(t--) solve();
}
 
 
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