无向图的最小环问题(Minimum Cycle) 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;
    const long long INF = (1LL << 58);
    vector<vector<long long>> g(n + 1, vector<long long>(n + 1, INF));
    for (int i = 1; i <= n; ++i) g[i][i] = 0;
    for (int i = 0; i < m; ++i) {
        int u, v;
        long long d;
        cin >> u >> v >> d;
        if (u != v) {
            g[u][v] = min(g[u][v], d);
            g[v][u] = min(g[v][u], d);
        }
    }
    vector<vector<long long>> dist = g;
    long long ans = INF;
    for (int k = 1; k <= n; ++k) {
        for (int i = 1; i < k; ++i) {
            if (g[i][k] >= INF) continue;
            for (int j = i + 1; j < k; ++j) {
                if (dist[i][j] < INF && g[k][j] < INF) {
                    ans = min(ans, dist[i][j] + g[i][k] + g[k][j]);
                }
            }
        }
        for (int i = 1; i <= n; ++i) {
            if (dist[i][k] >= INF) continue;
            for (int j = 1; j <= n; ++j) {
                if (dist[k][j] < INF) {
                    dist[i][j] = min(dist[i][j], dist[i][k] + dist[k][j]);
                }
            }
        }
    }
    if (ans >= INF) cout << "No solution.\n";
    else cout << ans << "\n";
    return 0;
}