結果

問題 No.386 貪欲な領主
ユーザー rpy3cpprpy3cpp
提出日時 2017-10-11 02:21:25
言語 C++14
(gcc 12.3.0 + boost 1.83.0)
結果
AC  
実行時間 299 ms / 2,000 ms
コード長 4,607 bytes
コンパイル時間 2,509 ms
コンパイル使用メモリ 182,696 KB
実行使用メモリ 39,224 KB
最終ジャッジ日時 2024-11-17 08:49:15
合計ジャッジ時間 4,837 ms
ジャッジサーバーID
（参考情報） 		judge1 / judge5
このコードへのチャレンジ
（要ログイン）

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 2 ms
5,248 KB
testcase_01 AC 2 ms
5,248 KB
testcase_02 AC 2 ms
5,248 KB
testcase_03 AC 2 ms
5,248 KB
testcase_04 AC 299 ms
39,224 KB
testcase_05 AC 272 ms
31,452 KB
testcase_06 AC 279 ms
31,360 KB
testcase_07 AC 4 ms
5,248 KB
testcase_08 AC 33 ms
6,016 KB
testcase_09 AC 6 ms
5,248 KB
testcase_10 AC 2 ms
5,248 KB
testcase_11 AC 2 ms
5,248 KB
testcase_12 AC 3 ms
5,248 KB
testcase_13 AC 9 ms
5,248 KB
testcase_14 AC 279 ms
31,376 KB
testcase_15 AC 258 ms
38,964 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 # 

#include <bits/stdc++.h>
using namespace std;
using vi = vector<int>;
using vvi = vector<vector<int>>;


class RMQidx{
private:
    int n;  // size of val
    vector<vector<int>> idx; // doubling range min index. idx[k][p]: index of min(val[p:p+2^k]) including p and not including p+2^k.
    vector<int> val;
    void init(const vector<int> &src){
        n = src.size();
        val = src;
        idx.emplace_back(vector<int>(n, 0));
        for (int i = 0; i != n; ++i) idx[0][i] = i;
        for (int k = 0, r = 1; r < n; ++k, r <<= 1){
            idx.emplace_back(idx[k]);
            for (int p = 0; p + r < n; ++p){
                auto idxL = idx[k][p];
                auto idxR = idx[k][p + r];
                idx[k + 1][p] = (val[idxL] > val[idxR]) ? idxR : idxL;
            }
        }
    }
public:
    RMQidx(): n(0), idx(vector<vector<int>>()), val(vector<int>()) {}
    RMQidx(const vector<int> &src){init(src);}
    int query(int L, int R){ // index of min(data[0][L..R] including both ends. [L, R]
        assert(L <= R);
        if (L == R) return L;
        int k = 31 - __builtin_clz(R - L);
        auto idxL = idx[k][L];
        auto idxR = idx[k][R + 1 - (1 << k)];
        return (val[idxL] > val[idxR])? idxR: idxL;
    }
};

class LCA{
private:
    int n;          // number of vertexes
    int root;
    const vector<vector<int>> & Es;
    vector<int> height;     // height[v] = height of vertex v.
    vector<int> i2v;        // i2v[i] = v. Euler tour order of vertexes.
    vector<int> v2i;        // v2i[v] = i. Last position of vertex v in the Euler tour.
    RMQidx rmq;             // Range Minimum Query object which returns the position (idx) of the minimum value in the euler tour.
    void dfs(int v, int h){
        height[v] = h;
        i2v.push_back(v);
        for (auto u : Es[v]){
            if (height[u] == -1) {
                dfs(u, h + 1);
                i2v.push_back(v);
            }
        }
    }
public:
    LCA(const vector<vector<int>> &_Es, int _root = 0):Es(_Es){
        n = Es.size();
        root = _root;
        height = vector<int>(n, -1);
        v2i = vector<int>(n, -1);
        dfs(root, 0);
        assert(i2v.size() == 2 * n - 1);
        vector<int> val(2 * n - 1);
        for (int i = 0; i != 2 * n - 1; ++i) {
            val[i] = height[i2v[i]];
            v2i[i2v[i]] = i;
        }
        rmq = RMQidx(val);
    }
    int query(int u, int v){
        // returns lowest common ancestor of vertex u and vertex v.
        int i = v2i[u];
        int j = v2i[v];
        if (i > j) swap(i, j);
        return i2v[rmq.query(i, j)];
    }
    int dist(int u, int v){
        // returns the distance between vertex u and vertex v.
        int k = query(u, v);
        return height[u] + height[v] - 2 * height[k];
    }
};

void dfs_et(int v, int p, vi & euler_tour, const vvi & Es){
    // v: current vertex,  p: parent of v
    euler_tour.push_back(v);
    for (auto u : Es[v]){
        if (u == p) continue;
        dfs_et(u, v, euler_tour, Es);
        euler_tour.push_back(v);
    }
}

vi get_euler_tour(const vvi &Es, int root = 0){
    vi euler_tour;
    int nV = Es.size();
    dfs_et(root, -1, euler_tour, Es);
    return euler_tour;
}

vi get_first_pos(const vi & et, int N){
    vi first(N, -1);
    for (int i = 0; i < et.size(); ++i){
        if (first[et[i]] == -1) first[et[i]] = i;
    }
    return first;
}

vi get_last_pos(const vi & et, int N){
    vi last(N, -1);
    for (int i = 0; i < et.size(); ++i){
        last[et[i]] = i;
    }
    return last;
}


int main() {
    int N;
    cin >> N;
    vector<vector<int>> Es(N);
    for (int i = 1; i < N; ++i){
        int a, b;
        cin >> a >> b;
        Es[a].push_back(b);
        Es[b].push_back(a);
    }
    LCA lca(Es);

    vi Us(N);
    for (auto & u : Us) cin >> u;

    auto euler_tour = get_euler_tour(Es);
    auto first = get_first_pos(euler_tour, N);
    auto last = get_last_pos(euler_tour, N);
    vector<long long> imos(2*N - 1, 0);

    int M;
    cin >> M;
    while (M--){
        int a, b, c;
        cin >> a >> b >> c;
        int d = lca.query(a, b);
        imos[first[a]] += c;
        imos[first[b]] += c;
        if (d != 0) imos[first[d] - 1] -= c;
        imos[first[d]] -= c;
    }
    vector<long long> cum(2 * N - 1, 0);
    cum[0] = imos[0];
    for (int i = 1; i < cum.size(); ++i){
        cum[i] = cum[i - 1] + imos[i];
    }
    long long cost = 0;
    cost += Us[0] * cum[last[0]];
    for (int i = 1; i < N; ++i){
        cost += Us[i] * (cum[last[i]] - cum[first[i] - 1]);
    }
    cout << cost;
    return 0;
}
0