题解：AT_abc404_c [ABC404C] Cycle Graph?

_peter6 · 2025-12-04 11:12:18 · 题解

解释题意

判断从任意一点出发把所有点都遍历一遍并且 n = m 。

首先可以发现如果 n \not = m，直接输出 No。

直接跑 dfs。

// code 1
#include <bits/stdc++.h>
using namespace std;

const int N = 2000010;
int ans[1010];
vector<vector<int> > vec(N);
bool vis[N];

void dfs(int x, int fa) {
    vis[x] = 1;
    for (auto u : vec[x]) {
        if (u != fa && !vis[u]) {
            dfs(u, x);
        }
    }
}

int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int n, m;
    cin >> n >> m;
    if (n != m) {
        cout << "No";
        return 0;
    }
    for (int i = 1; i <= m; i++) {
        int u, v;
        cin >> u >> v;
        vec[u].push_back(v);
        vec[v].push_back(u);
    }
    for (int i = 1; i <= n; i++) {
        if (vec[i].size() != 2) {
            cout << "No";
            return 0;
        }
    }
    dfs(1, 0);
    for (int i = 1; i <= n; i++) {
        if (!vis[i]) {
            cout << "No";
            return 0;
        }
    }
    cout << "Yes";
    return 0;
}

并查集写法

// code 2
#include<bits/stdc++.h>
using namespace std;

const int N = 2000010;
int fa[N];
void init(int n) {
    for (int i = 1; i <= n; i++)  {
        fa[i] = i;
    }
}
int find(int x) {
    if(fa[x] != x) fa[x] = find(fa[x]);
    return fa[x];
}
void merge(int x,int y) {
    x = find(x); y = find(y);
    if (x != y) {
        fa[x] = y;
    }
}

vector<int> degree(N);

int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int n, m;
    cin >> n >> m;
    init(n);
    vector<pair<int, int> > edge;
    for (int i = 0; i < m; i++) {
        int u, v;
        cin >> u >> v;
        edge.push_back({u, v});
        degree[u]++;
        degree[v]++;
    }
    for (int i = 1; i <= n; i++)  {
        if (degree[i] != 2) {
            cout << "No" << endl;
            return 0;
        }
    }
    int cnt=0;
    for (int i = 0; i < m; i++) {
        int a = edge[i].first;
        int b = edge[i].second;
        if (find(a) == find(b)) {
            cnt++;
        } 
        else {
            merge(a, b);
        }
    }
    int root = find(1);
    bool flag = 1;
    for (int i = 2; i <= n; i++) {
        if (find(i) != root) {
            flag = 0;
            break;
        }
    }
    if (cnt == 1 && m == n && flag)  {
        cout << "Yes" << endl;
    } 
    else {
        cout << "No" << endl;
    }
    return 0;
}