結果

問題 No.1212 Second Path
ユーザー SSRS
提出日時 2020-08-30 17:02:27
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 2,116 ms / 3,000 ms
コード長 4,236 bytes
コンパイル時間 3,160 ms
コンパイル使用メモリ 229,212 KB
最終ジャッジ日時 2025-01-14 00:24:08
ジャッジサーバーID
（参考情報） 		judge3 / judge4
このコードへのチャレンジ
（要ログイン）
ファイルパターン 結果
sample AC * 3
other AC * 45
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;
const int INF = 1000000000;
const int LOG = 18;
struct lowest_common_ancestor{
	vector<int> d;
	vector<vector<int>> p;
	lowest_common_ancestor(){
	}
	lowest_common_ancestor(vector<int> &P, vector<vector<int>> &C){
		int N = P.size();
		d = vector<int>(N, 0);
		queue<int> Q;
		Q.push(0);
		while (!Q.empty()){
			int v = Q.front();
			Q.pop();
			for (int w : C[v]){
				d[w] = d[v] + 1;
				Q.push(w);
			}
		}
		p = vector<vector<int>>(LOG, vector<int>(N, -1));
		for (int i = 0; i < N; i++){
			p[0][i] = P[i];
		}
		for (int i = 1; i < LOG; i++){
			for (int j = 0; j < N; j++){
				if (p[i - 1][j] != -1){
					p[i][j] = p[i - 1][p[i - 1][j]];
				}
			}
		}
	}
	int query(int u, int v){
		if (d[u] > d[v]){
			swap(u, v);
		}
		for (int k = 0; k < LOG; k++){
			if ((d[v] - d[u]) >> k & 1){
				v = p[k][v];
			}
		}
		if (u == v){
		    return u;
		}
		for (int k = LOG - 1; k >= 0; k--){
			if (p[k][u] != p[k][v]){
				u = p[k][u];
				v = p[k][v];
    		}
		}
		return p[0][u];
	}
};
template <typename T>
struct binary_indexed_tree{
	int N;
	vector<T> BIT;
	binary_indexed_tree(){
	}
	binary_indexed_tree(int n){
		N = 1;
		while (N < n){
			N *= 2;
		}
		BIT = vector<T>(N + 1, 0);
	}
	void add(int i, T x){
		i++;
		while (i <= N){
			BIT[i] += x;
			i += i & -i;
		}
	}
	T sum(int i){
		T ans = 0;
		while (i > 0){
			ans += BIT[i];
			i -= i & -i;
		}
		return ans;
	}
	T query(int L, int R){
		return sum(R) - sum(L);
	}
};
template <typename T>
struct euler_tour{
	vector<T> A;
	vector<int> left;
	vector<int> right;
	binary_indexed_tree<T> BIT;
	void dfs(vector<vector<int>> &c, int v){
		left[v] = A.size();
		A.push_back(0);
		for (int w : c[v]){
			dfs(c, w);
		}
		right[v] = A.size();
		A.push_back(0);
	}
	euler_tour(vector<int> &p, vector<vector<int>> &c){
		int N = p.size();
		left = vector<int>(N);
		right = vector<int>(N);
		dfs(c, 0);
		BIT = binary_indexed_tree<int>(N * 2);
	}
	void add(int v){
		BIT.add(left[v], 1);
		BIT.add(right[v], -1);
	}
	T query(int v, int w, int u){
		return BIT.query(left[u], left[v] + 1) + BIT.query(left[u] + 1, left[w] + 1);
	}
};
int main(){
  int N;
  cin >> N;
  vector<vector<pair<long long, int>>> E(N);
  vector<tuple<int, int, int>> edges;
  for (int i = 0; i < N - 1; i++){
    int u, v, w;
    cin >> u >> v >> w;
    u--;
    v--;
    E[u].push_back(make_pair(w, v));
    E[v].push_back(make_pair(w, u));
    edges.push_back(make_tuple(w, u, v));
  }
  vector<int> p(N, -1);
  vector<vector<int>> c(N);
  vector<long long> s(N, 0);
  queue<int> q;
  q.push(0);
  while (!q.empty()){
    int v = q.front();
    q.pop();
    for (auto e : E[v]){
      int d = e.first;
      int w = e.second;
      if (w != p[v]){
        p[w] = v;
        c[v].push_back(w);
        s[w] = s[v] + d;
        q.push(w);
      }
    }
  }
  lowest_common_ancestor LCA(p, c);
  int Q;
  cin >> Q;
  vector<int> x(Q), y(Q), l(Q);
  for (int i = 0; i < Q; i++){
    cin >> x[i] >> y[i];
    x[i]--;
    y[i]--;
    l[i] = LCA.query(x[i], y[i]);
  }
  sort(edges.begin(), edges.end());
  vector<int> tv(Q, N);
  vector<int> fv(Q, 0);
  while (1){
    bool upd = false;
    vector<vector<int>> check(N - 1);
    for (int i = 0; i < Q; i++){
      if (tv[i] - fv[i] > 1){
        upd = true;
        int mid = (tv[i] + fv[i]) / 2;
        check[mid - 1].push_back(i);
      }
    }
    if (!upd){
      break;
    }
    euler_tour<int> T1(p, c), T2(p, c);
    vector<bool> used(N, false);
    for (int i = 0; i < N - 1; i++){
      int u = get<1>(edges[i]);
      int v = get<2>(edges[i]);
      T1.add(u);
      T1.add(v);
      if (u == p[v]){
        T2.add(v);
        used[v] = true;
      } else {
        T2.add(u);
        used[u] = true;
      }
      for (int id : check[i]){
        int a = T1.query(x[id], y[id], l[id]);
        int b = T2.query(x[id], y[id], l[id]);
        if (used[l[id]]){
          b--;
        }
        if (a > b * 2){
          tv[id] = i + 1;
        } else {
          fv[id] = i + 1;
        }
      }
    }
  }
  for (int i = 0; i < Q; i++){
    if (tv[i] == N){
      cout << -1 << endl;
    } else {
      cout << s[x[i]] + s[y[i]] - s[l[i]] * 2 + get<0>(edges[tv[i] - 1]) * 2 << endl;
    }
  }
}
0