（特殊最短路）给定一个含 $N$ 个点、$M$ 条边的带权无向图，边权非负。起点为 $S$，终点为 $T$。对于一条 $S$ 到 $T$ 的路径，可以在整条路径中，至多选择一条边作为“免费边”：当第一次经过这条被选中的边时，费用视为 0；如果之后再次经过该边，则仍按其原始权重计费。点和边均允许重复经过。求从 $S$ 到 $T$ 的最小总费用。

以下代码求解了上述问题。试补全程序。

1#include <algorithm>2#include <iostream>3#include <queue>4#include <vector>5using namespace std;6 7const long long INF = 1e18;8 9struct Edge {10    int to;11    int weight;12};13 14struct State {15    long long dist;16    int u;17    int used_freebie; // 0 for not used, 1 for used18    bool operator>(const State &other) const {19        return dist > other.dist;20    }21};22 23int main() {24    int n, m, s, t;25    cin >> n >> m >> s >> t;26 27    vector<vector<Edge>> adj(n + 1);28    for (int i = 0; i < m; ++i) {29        int u, v, w;30        cin >> u >> v >> w;31        adj[u].push_back({v, w});32        adj[v].push_back({u, w});33    }34 35    vector<vector<long long>> d(n + 1, vector<long long>(2, INF));36    priority_queue<State, vector<State>, greater<State>> pq;37 38    d[s][0] = 0;39    pq.push({0, s, __①__});40 41    while (!pq.empty()) {42        State current = pq.top();43        pq.pop();44 45        long long dist = current.dist;46        int u = current.u;47        int used = current.used_freebie;48 49        if (dist > __②__) {50            continue;51        }52 53        for (const auto &edge : adj[u]) {54            int v = edge.to;55            int w = edge.weight;56 57            if (d[u][used] + w < __③__) {58                __③__ = d[u][used] + w;59                pq.push({__③__, v, used});60            }61 62            if (used == 0) {63                if (__④__ < d[v][1]) {64                    d[v][1] = d[u][used];65                    pq.push({d[v][1], v, 1});66                }67            }68        }69    }70 71    cout << __⑤__ << endl;72    return 0;73}