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给出经济路线以及商业路线#xff0c;在给出起始点s#xff0c;终止点e#xff0c;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#xff0c;则需要从s-u,u-v#xff0c;v-e点的路径最短。使用Dijkstra计算出从s点…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-vv-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-vv-e点中路线最短的
代码
#include bits/stdc.husing 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;vectorEdge edges;vectorint 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_queueHeapNode 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, dequeint 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;}DijkstraMAXN, 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;dequeint 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;
}
