結果

問題 No.898 tri-βutree
ユーザー fumiphysfumiphys
提出日時 2019-10-04 21:40:33
言語 C++14
(gcc 13.2.0 + boost 1.83.0)
結果
AC  
実行時間 523 ms / 4,000 ms
コード長 5,137 bytes
コンパイル時間 2,546 ms
コンパイル使用メモリ 185,064 KB
実行使用メモリ 38,032 KB
最終ジャッジ日時 2023-08-08 15:53:09
合計ジャッジ時間 12,658 ms
ジャッジサーバーID
（参考情報） 		judge15 / judge11
このコードへのチャレンジ（β）

テストケース

テストケース表示
入力 結果 実行時間
実行使用メモリ
testcase_00 AC 227 ms
38,032 KB
testcase_01 AC 3 ms
8,108 KB
testcase_02 AC 3 ms
8,052 KB
testcase_03 AC 3 ms
8,148 KB
testcase_04 AC 4 ms
8,184 KB
testcase_05 AC 4 ms
8,120 KB
testcase_06 AC 3 ms
8,064 KB
testcase_07 AC 512 ms
28,700 KB
testcase_08 AC 500 ms
28,440 KB
testcase_09 AC 523 ms
28,548 KB
testcase_10 AC 523 ms
28,528 KB
testcase_11 AC 498 ms
28,388 KB
testcase_12 AC 514 ms
28,424 KB
testcase_13 AC 509 ms
28,436 KB
testcase_14 AC 516 ms
28,400 KB
testcase_15 AC 503 ms
28,452 KB
testcase_16 AC 494 ms
28,392 KB
testcase_17 AC 507 ms
28,444 KB
testcase_18 AC 512 ms
28,540 KB
testcase_19 AC 504 ms
28,396 KB
testcase_20 AC 506 ms
28,516 KB
testcase_21 AC 503 ms
28,496 KB
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 # 

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

// macros
#define pb emplace_back
#define mk make_pair
#define FOR(i, a, b) for(int i=(a);i<(b);++i)
#define rep(i, n) FOR(i, 0, n)
#define rrep(i, n) for(int i=((int)(n)-1);i>=0;i--)
#define irep(itr, st) for(auto itr = (st).begin(); itr != (st).end(); ++itr)
#define irrep(itr, st) for(auto itr = (st).rbegin(); itr != (st).rend(); ++itr)
#define all(x) (x).begin(),(x).end()
#define sz(x) ((int)(x).size())
#define UNIQUE(v) v.erase(unique(v.begin(), v.end()), v.end())
#define bit(n) (1LL<<(n))
// functions
template <class T>bool chmax(T &a, const T &b){if(a < b){a = b; return 1;} return 0;}
template <class T>bool chmin(T &a, const T &b){if(a > b){a = b; return 1;} return 0;}
template <typename T> istream &operator>>(istream &is, vector<T> &vec){for(auto &v: vec)is >> v; return is;}
template <typename T> ostream &operator<<(ostream &os, const vector<T>& vec){for(int i = 0; i < vec.size(); i++){ os << vec[i]; if(i + 1 != vec.size())os << " ";} return os;}
template <typename T> ostream &operator<<(ostream &os, const set<T>& st){for(auto itr = st.begin(); itr != st.end(); ++itr){ os << *itr; auto titr = itr; if(++titr != st.end())os << " ";} return os;}
template <typename T> ostream &operator<<(ostream &os, const unordered_set<T>& st){for(auto itr = st.begin(); itr != st.end(); ++itr){ os << *itr; auto titr = itr; if(++titr != st.end())os << " ";} return os;}
template <typename T> ostream &operator<<(ostream &os, const multiset<T>& st){for(auto itr = st.begin(); itr != st.end(); ++itr){ os << *itr; auto titr = itr; if(++titr != st.end())os << " ";} return os;}
template <typename T> ostream &operator<<(ostream &os, const unordered_multiset<T>& st){for(auto itr = st.begin(); itr != st.end(); ++itr){ os << *itr; auto titr = itr; if(++titr != st.end())os << " ";} return os;}
template <typename T1, typename T2> ostream &operator<<(ostream &os, const pair<T1, T2> &p){os << p.first << " " << p.second; return os;}
template <typename T1, typename T2> ostream &operator<<(ostream &os, const map<T1, T2> &mp){for(auto itr = mp.begin(); itr != mp.end(); ++itr){ os << itr->first << ":" << itr->second; auto titr = itr; if(++titr != mp.end())os << " "; } return os;}
template <typename T1, typename T2> ostream &operator<<(ostream &os, const unordered_map<T1, T2> &mp){for(auto itr = mp.begin(); itr != mp.end(); ++itr){ os << itr->first << ":" << itr->second; auto titr = itr; if(++titr != mp.end())os << " "; } return os;}
//  types
using ll = long long int;
using P = pair<int, int>;
// constants
const int inf = 1e9;
const ll linf = 1LL << 50;
const double EPS = 1e-10;
const int mod = 1000000007;
const int dx[4] = {-1, 0, 1, 0};
const int dy[4] = {0, -1, 0, 1};
// io
struct fast_io{
  fast_io(){ios_base::sync_with_stdio(false); cin.tie(0); cout << fixed << setprecision(20);}
} fast_io_;

struct LCATree{
  int n = 0, m = 0;
  vector<vector<int>> edge;
  vector<vector<int>> par;
  vector<bool> vis;
  vector<int> h;
  LCATree(){}
  LCATree(int n): n(n){
    while((1LL<<m) <= n)m++;
    edge.resize(n);
    par.assign(m + 1, vector<int>(n, 0));
    vis.resize(n, false);
    h.resize(n);
  }
  void adde(int from, int to){
    edge[from].emplace_back(to);
  }
  void dfs(int i){
    vis[i] = true;
    for(auto e: edge[i]){
      if(!vis[e]){
        par[0][e] = i;
        h[e] = h[i] + 1;
        dfs(e);
      }
    }
  }
  void build(){
    fill(vis.begin(), vis.end(), false);
    fill(h.begin(), h.end(), 0);
    par[0][0] = 0;
    h[0] = 0;
    dfs(0);
    for(int i = 1; i <= m; i++){
      for(int j = 0; j < n; j++){
        par[i][j] = par[i-1][par[i-1][j]];
      }
    }
  }
  int go_up(int u, int k){
    int i = 0;
    while(k){
      if(k % 2 == 1){
        u = par[i][u];
      }
      i++;
      k /= 2;
    }
    return u;
  }
  int lca(int u, int v){
    if(h[u] > h[v])u = go_up(u, h[u] - h[v]);
    if(h[u] < h[v])v = go_up(v, h[v] - h[u]);
    if(u == v)return u;
    int ld = 0, rd = n;
    while(rd - ld > 1){
      int k = (rd + ld) / 2;
      int uk = go_up(u, k);
      int vk = go_up(v, k);
      if(uk == vk)rd = k;
      else ld = k;
    }
    return go_up(u, rd);
  }
};


vector<pair<int, ll>> edge[100010], child[100010];
bool vis[100010];
ll val[100010], pre[100010];

void dfs(int i){
  vis[i] = true;
  for(auto e: edge[i]){
    if(vis[e.first])continue;
    child[i].pb(e);
    val[e.first] = val[i] + e.second;
    pre[e.first] = val[i];
    dfs(e.first);
  }
}

int main(int argc, char const* argv[])
{
  int n; cin >> n;
  rep(i, n - 1){
    int u, v; cin >> u >> v;
    ll w; cin >> w;
    edge[u].pb(v, w);
    edge[v].pb(u, w);
  }
  dfs(0);
  LCATree t(n);
  rep(i, n){
    for(auto e: child[i])t.adde(i, e.first);
  }
  t.build();
  int q; cin >> q;
  rep(i, q){
    int x, y, z; cin >> x >> y >> z;
    int lca = t.lca(x, t.lca(y, z));
    ll res = 0;
    res += val[x] - val[lca];
    res += val[y] - val[lca];
    res += val[z] - val[lca];
    res -= val[t.lca(x, y)] - val[lca];
    res -= val[t.lca(y, z)] - val[lca];
    res -= val[t.lca(z, x)] - val[lca];
    cout << res << endl;
  }
  return 0;
}
0