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题意
给出经济路线以及商业路线，在给出起始点s，终止点e，在只能使用其中一个商业路线 的情况下输出最短路径
思路
如果选择商业路线为从u到v，则需要从s->u,u->v，v->e点的路径最短。使用Dijkstra计算出从s点到其它各点，以及从e点到其它各点的最短路径，然后遍历商业路线u,v，选取从s->u,u->v，v->e点中路线最短的
代码
#include <bits/stdc++.h>using namespace std;#define _for(i, a, b) for(int i = (a); i < (b); i++)
#define _rep(i, a, b) for (int i = (a); i <= (b); i++)struct Edge
{int u, v, d;
};struct HeapNode
{int u, d;bool operator<(const HeapNode& other) const{return d > other.d;}
};template <int SZV, int INF>
struct Dijkstra
{int n;vector<Edge> edges;vector<int> graph[SZV];bool done[SZV];int d[SZV], p[SZV];void init(int n){this->n = n;edges.clear();_for(i, 0, n) {graph[i].clear();}}void addEdge(int u, int v, int d){graph[u].push_back(edges.size());edges.push_back({u, v, d});}void dijkstra(int s){priority_queue<HeapNode> pq;fill_n(done, n, false);fill_n(d, n, INF);d[s] = 0;pq.push({s, 0});while (!pq.empty()) {HeapNode curNode = pq.top();pq.pop();int u = curNode.u;if (done[u]) {continue;}done[u] = true;_for(i, 0, graph[u].size()) {const auto& edge = edges[graph[u][i]];int v = edge.v;if (d[u] + edge.d < d[v]) {d[v] = d[u] + edge.d;p[v] = graph[u][i];pq.push({v, d[v]});}}}}void getPath(int s, int e, deque<int>& path, bool rev = false){int x = e;if (rev) {path.push_back(x);} else {path.push_front(x);}while (x != s) {x = edges[p[x]].u;if (rev) {path.push_back(x);} else {path.push_front(x);}}}
};void fastio()
{ios_base::sync_with_stdio(false);cin.tie(nullptr);cout.tie(nullptr);
}const int MAXN = 500 + 4;
const int INF = 1e9;int main()
{fastio();#ifndef ONLINE_JUDGEifstream fin("f:\\OJ\\uva_in.txt");streambuf* back = cin.rdbuf(fin.rdbuf());#endifint N, S, E;int kase = 0;while (cin >> N >> S >> E) {if (kase++) {cout << endl;}Dijkstra<MAXN, INF> sd, ed;sd.init(N + 1); ed.init(N + 1);int M;cin >> M;_for(i, 0, M) {int X, Y, Z;cin >> X >> Y >> Z;sd.addEdge(X, Y, Z);sd.addEdge(Y, X, Z);ed.addEdge(X, Y, Z);ed.addEdge(Y, X, Z);}sd.dijkstra(S);ed.dijkstra(E);int cu = -1;int ans = INF;deque<int> path;if (sd.d[E] < ans) {ans = sd.d[E];sd.getPath(S, E, path);}auto update = [&](int u, int v, int d) {if (sd.d[u] < ans && ed.d[v] < ans && sd.d[u] + d + ed.d[v] < ans) {ans = sd.d[u] + d + ed.d[v];cu = u;path.clear();sd.getPath(S, u, path);ed.getPath(E, v, path, true);}};int K;cin >> K;_for(i, 0, K) {int u, v, d;cin >> u >> v >> d;update(u, v, d);update(v, u, d);}_for(i, 0, path.size()) {if (i) {cout << " ";}cout << path[i];}cout << endl;if (cu == -1) {cout << "Ticket Not Used" << endl;} else {cout << cu << endl;}cout << ans << endl;}#ifndef ONLINE_JUDGEcin.rdbuf(back);#endifreturn 0;
}