视频推荐(MooTube G) 题解

思路分析

按题目要求使用对应数据结构与算法模拟/递推求解，注意数据范围边界与时间复杂度。

参考代码

#include <iostream>
#include <vector>
#include <numeric>
#include <algorithm>
using namespace std;

struct Edge {
    int u, v;
    long long w;
    bool operator<(const Edge& o) const { return w > o.w; }
};

struct Query {
    long long k;
    int v, id;
    bool operator<(const Query& o) const { return k > o.k; }
};

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int n, q;
    if (!(cin >> n >> q)) return 0;
    vector<Edge> edges(n - 1);
    for (int i = 0; i < n - 1; ++i) {
        cin >> edges[i].u >> edges[i].v >> edges[i].w;
    }
    sort(edges.begin(), edges.end());
    vector<Query> qs(q);
    for (int i = 0; i < q; ++i) {
        cin >> qs[i].k >> qs[i].v;
        qs[i].id = i;
    }
    sort(qs.begin(), qs.end());
    vector<int> p(n + 1), sz(n + 1, 1), ans(q);
    iota(p.begin(), p.end(), 0);
    auto find = [&](int x) {
        while (p[x] != x) x = p[x] = p[p[x]];
        return x;
    };
    int eIdx = 0;
    for (int i = 0; i < q; ++i) {
        while (eIdx < n - 1 && edges[eIdx].w >= qs[i].k) {
            int ru = find(edges[eIdx].u), rv = find(edges[eIdx].v);
            if (ru != rv) {
                p[ru] = rv;
                sz[rv] += sz[ru];
            }
            ++eIdx;
        }
        ans[qs[i].id] = sz[find(qs[i].v)] - 1;
    }
    for (int i = 0; i < q; ++i) cout << ans[i] << "\n";
    return 0;
}