题解

解题

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

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);

    int N, M;
    cin >> N >> M;

    // 存储每个观察结果对应的边
    vector<vector<pair<int, int>>> edges_by_obs;

    for (int i = 0; i < M; ++i) {
        int m;
        cin >> m;
        vector<int> cows(m);
        for (int j = 0; j < m; ++j) {
            cin >> cows[j];
        }
        // 生成连续对的边
        vector<pair<int, int>> edges;
        for (int j = 0; j < m - 1; ++j) {
            edges.emplace_back(cows[j], cows[j + 1]);
        }
        edges_by_obs.push_back(edges);
    }

    // 二分查找最大的X
    int low = 0, high = M, best_X = 0;
    while (low <= high) {
        int mid = (low + high) / 2;

        // 构建图
        vector<vector<int>> adj(N + 1);
        vector<int> in_degree(N + 1, 0);
        for (int i = 0; i < mid; ++i) {
            for (auto& e : edges_by_obs[i]) {
                int u = e.first, v = e.second;
                adj[u].push_back(v);
                in_degree[v]++;
            }
        }

        // Kahn's算法检查是否为DAG
        queue<int> q;
        vector<int> temp_in = in_degree;
        for (int i = 1; i <= N; ++i) {
            if (temp_in[i] == 0) {
                q.push(i);
            }
        }

        int cnt = 0;
        while (!q.empty()) {
            int u = q.front();
            q.pop();
            cnt++;
            for (int v : adj[u]) {
                temp_in[v]--;
                if (temp_in[v] == 0) {
                    q.push(v);
                }
            }
        }

        if (cnt == N) {
            best_X = mid;
            low = mid + 1;
        } else {
            high = mid - 1;
        }
    }

    // 生成最佳X对应的字典序最小拓扑排序
    vector<vector<int>> adj_final(N + 1);
    vector<int> in_degree_final(N + 1, 0);
    for (int i = 0; i < best_X; ++i) {
        for (auto& e : edges_by_obs[i]) {
            int u = e.first, v = e.second;
            adj_final[u].push_back(v);
            in_degree_final[v]++;
        }
    }

    // 使用最小堆确保字典序最小
    priority_queue<int, vector<int>, greater<int>> pq;
    for (int i = 1; i <= N; ++i) {
        if (in_degree_final[i] == 0) {
            pq.push(i);
        }
    }

    vector<int> result;
    while (!pq.empty()) {
        int u = pq.top();
        pq.pop();
        result.push_back(u);
        for (int v : adj_final[u]) {
            in_degree_final[v]--;
            if (in_degree_final[v] == 0) {
                pq.push(v);
            }
        }
    }

    // 输出结果
    for (int i = 0; i < N; ++i) {
        if (i > 0) cout << " ";
        cout << result[i];
    }
    cout << endl;

    return 0;
}
   

看到这了，给个赞呗