英国菜单(British Menu) 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 <algorithm>
using namespace std;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int n, m;
    if (!(cin >> n >> m)) return 0;
    vector<vector<int>> adj(n + 1);
    for (int i = 0; i < m; ++i) {
        int u, v;
        cin >> u >> v;
        adj[u].push_back(v);
    }
    vector<int> dfn(n + 1, 0), low(n + 1, 0), in_stk(n + 1, 0), scc(n + 1, 0);
    vector<int> stk, call_u, call_idx;
    vector<vector<int>> scc_nodes(1);
    int timer = 0, scc_cnt = 0;
    for (int i = 1; i <= n; ++i) {
        if (!dfn[i]) {
            call_u.push_back(i);
            call_idx.push_back(0);
            dfn[i] = low[i] = ++timer;
            stk.push_back(i);
            in_stk[i] = 1;
            while (!call_u.empty()) {
                int u = call_u.back();
                int& idx = call_idx.back();
                if (idx < (int)adj[u].size()) {
                    int v = adj[u][idx++];
                    if (!dfn[v]) {
                        dfn[v] = low[v] = ++timer;
                        stk.push_back(v);
                        in_stk[v] = 1;
                        call_u.push_back(v);
                        call_idx.push_back(0);
                    } else if (in_stk[v]) {
                        low[u] = min(low[u], dfn[v]);
                    }
                } else {
                    if (low[u] == dfn[u]) {
                        ++scc_cnt;
                        vector<int> comp;
                        while (true) {
                            int x = stk.back();
                            stk.pop_back();
                            in_stk[x] = 0;
                            scc[x] = scc_cnt;
                            comp.push_back(x);
                            if (x == u) break;
                        }
                        scc_nodes.push_back(comp);
                    }
                    call_u.pop_back();
                    call_idx.pop_back();
                    if (!call_u.empty()) {
                        low[call_u.back()] = min(low[call_u.back()], low[u]);
                    }
                }
            }
        }
    }
    vector<int> dp(n + 1, 1), best_out(n + 1, 0), vis(n + 1, 0);
    int ans = 1;
    for (int c = 1; c <= scc_cnt; ++c) {
        const auto& comp = scc_nodes[c];
        for (int u : comp) {
            int bo = 0;
            for (int v : adj[u]) {
                if (scc[v] != c) bo = max(bo, dp[v]);
            }
            best_out[u] = bo;
        }
        auto dfs_comp = [&](auto& self, int u, int len) -> int {
            int res = len + best_out[u];
            vis[u] = 1;
            for (int v : adj[u]) {
                if (scc[v] == c && !vis[v]) {
                    res = max(res, self(self, v, len + 1));
                }
            }
            vis[u] = 0;
            return res;
        };
        for (int u : comp) {
            dp[u] = dfs_comp(dfs_comp, u, 1);
            ans = max(ans, dp[u]);
        }
    }
    cout << ans << "\n";
    return 0;
}