排队方案(Queueing Arrangements) 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 <algorithm>
#include <iostream>
#include <numeric>
#include <utility>
#include <vector>
using namespace std;

const int MOD = 1000000007;

struct DSU {
    vector<int> p;
    DSU(int n) : p(n + 1) { iota(p.begin(), p.end(), 0); }
    int find(int x) { return p[x] == x ? x : p[x] = find(p[x]); }
    bool unite(int a, int b) {
        a = find(a);
        b = find(b);
        if (a == b) return false;
        p[a] = b;
        return true;
    }
};

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int n, m;
    if (!(cin >> n >> m)) return 0;
    vector<pair<int, int>> edges(m);
    for (int i = 0; i < m; ++i) cin >> edges[i].first >> edges[i].second;
    sort(edges.begin(), edges.end());
    edges.erase(unique(edges.begin(), edges.end()), edges.end());

    vector<int> indeg(n + 1, 0), outdeg(n + 1, 0);
    DSU dsu(n);
    int comps = n;
    for (auto [u, v] : edges) {
        if (u == v) {
            cout << 0 << "\n";
            return 0;
        }
        outdeg[u]++;
        indeg[v]++;
        if (outdeg[u] > 1 || indeg[v] > 1) {
            cout << 0 << "\n";
            return 0;
        }
        if (!dsu.unite(u, v)) {
            cout << 0 << "\n";
            return 0;
        }
        comps--;
    }
    long long ans = 1;
    for (int i = 1; i <= comps; ++i) ans = ans * i % MOD;
    cout << ans << "\n";
    return 0;
}