解法一：Dijkstra

#include <bits/stdc++.h>
using namespace std;
const int N = 505;
const int INF = 0x3f3f3f3f;
class Edge {
public:
    int v;
    int dis;
    int cost;
    Edge() {}
    Edge(int v, int dis, int cost) : v(v), dis(dis), cost(cost) {}
};
vector<Edge> graph[N];
bool vis[N];
int dis[N];
int cost[N];
int pre[N];
int n, m, s, d;
void dijkstra(int s, int d) {
    memset(vis, false, sizeof(vis));
    fill(dis, dis + N, INF);
    fill(cost, cost + N, INF);
    dis[s] = 0;
    cost[s] = 0;
    while (!vis[d]) {
        int min = 0x3f3f3f3f;
        int u = -1;
        for (int i = 0; i < n; ++i) {
            if (!vis[i] && dis[i] < min) {
                u = i;
                min = dis[i];
            }
        }
        if (u == -1) {
            break;
        }
        vis[u] = true;
        for (Edge edge:graph[u]) {
            if ((dis[u] + edge.dis < dis[edge.v]) ||
                ((dis[u] + edge.dis == dis[edge.v]) && (cost[u] + edge.cost < cost[edge.v]))) {
                dis[edge.v] = dis[u] + edge.dis;
                pre[edge.v] = u;
                cost[edge.v] = cost[u] + edge.cost;
            }
        }
    }
}
void printPath(int s, int d) {
    if (s == d) {
        cout << d << " ";
        return;
    }
    printPath(s, pre[d]);
    cout << d << " ";
}
int main() {
    ios::sync_with_stdio(false);
    cin >> n >> m >> s >> d;
    int u, v, _dis, _cost;
    for (int i = 0; i < m; ++i) {
        cin >> u >> v >> _dis >> _cost;
        graph[u].emplace_back(Edge(v, _dis, _cost));
        graph[v].emplace_back(Edge(u, _dis, _cost));
    }
    dijkstra(s, d);
    printPath(s, d);
    cout << dis[d] << " " << cost[d] << '\n';
}