唯一漫步(Unique Walk) 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>
using namespace std;

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 + 1);
    for (int i = 1; i <= m; ++i) {
        cin >> edges[i].first >> edges[i].second;
    }
    int k;
    cin >> k;
    vector<int> in_s(m + 1, 0), s_list(k);
    for (int i = 0; i < k; ++i) {
        cin >> s_list[i];
        in_s[s_list[i]] = 1;
    }
    vector<int> p(n + 1);
    iota(p.begin(), p.end(), 0);
    auto find = [&](int x) {
        while (p[x] != x) x = p[x] = p[p[x]];
        return x;
    };
    for (int i = 1; i <= m; ++i) {
        if (!in_s[i]) {
            int ru = find(edges[i].first), rv = find(edges[i].second);
            if (ru != rv) p[ru] = rv;
        }
    }
    vector<int> deg(n + 1, 0), p2(n + 1);
    iota(p2.begin(), p2.end(), 0);
    auto find2 = [&](int x) {
        while (p2[x] != x) x = p2[x] = p2[p2[x]];
        return x;
    };
    for (int idx : s_list) {
        int u = find(edges[idx].first);
        int v = find(edges[idx].second);
        deg[u]++;
        deg[v]++;
        int ru = find2(u), rv = find2(v);
        if (ru != rv) p2[ru] = rv;
    }
    int root = find2(find(edges[s_list[0]].first));
    bool connected = true;
    for (int idx : s_list) {
        int u = find(edges[idx].first);
        if (find2(u) != root) {
            connected = false;
            break;
        }
    }
    int odd = 0;
    for (int i = 1; i <= n; ++i) {
        if (p[i] == i && (deg[i] % 2 == 1)) {
            odd++;
        }
    }
    if (connected && (odd == 0 || odd == 2)) {
        cout << "Yes\n";
    } else {
        cout << "No\n";
    }
    return 0;
}