稳定婚姻(Stable Marriage) 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 <string>
#include <unordered_map>
#include <algorithm>
using namespace std;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int n;
    if (!(cin >> n)) return 0;
    unordered_map<string, int> id_map;
    id_map.reserve(2 * n + 7);
    int V = 2 * n;
    vector<vector<int>> adj(V + 1);
    for (int i = 1; i <= n; ++i) {
        string g, b;
        cin >> g >> b;
        id_map[g] = i;
        id_map[b] = n + i;
        adj[i].push_back(n + i);
    }
    int m;
    cin >> m;
    for (int i = 0; i < m; ++i) {
        string g, b;
        cin >> g >> b;
        int gid = id_map[g];
        int bid = id_map[b];
        adj[bid].push_back(gid);
    }
    vector<int> dfn(V + 1, 0), low(V + 1, 0), in_stk(V + 1, 0), scc(V + 1, 0);
    vector<int> stk, call_u, call_idx;
    int timer = 0, scc_cnt = 0;
    for (int i = 1; i <= V; ++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;
                        while (true) {
                            int x = stk.back();
                            stk.pop_back();
                            in_stk[x] = 0;
                            scc[x] = scc_cnt;
                            if (x == u) break;
                        }
                    }
                    call_u.pop_back();
                    call_idx.pop_back();
                    if (!call_u.empty()) {
                        low[call_u.back()] = min(low[call_u.back()], low[u]);
                    }
                }
            }
        }
    }
    for (int i = 1; i <= n; ++i) {
        cout << (scc[i] == scc[n + i] ? "Unsafe\n" : "Safe\n");
    }
    return 0;
}