Ultraviolet A 1048-low cost air travel (Shortest Path)

Ultraviolet A 1048-low cost air travel

Question Link

Question: given some tickets, the minimum cost of these trips is required for certain itineraries.

Idea: the shortest path: the number of edges in several cities corresponding to a ticket. The Node indicates that I is formed at the moment and the status of city J is formed. In this way, the shortest path is implemented, note that there are pitfalls in this question, that is, the city number may be large, so various hash

Code:

#include <cstdio>#include <cstring>#include <vector>#include <queue>#include <map>using namespace std;const int MAXNODE = 50005;const int MAXEDGE = 1000005;typedef int Type;const Type INF = 0x3f3f3f3f;struct Edge {int u, v, id;Type dist;Edge() {}Edge(int u, int v, Type dist, int id) {this->u = u;this->v = v;this->dist = dist;this->id = id;}};struct HeapNode {Type d;int u;HeapNode() {}HeapNode(Type d, int u) {this->d = d;this->u = u;}bool operator < (const HeapNode& c) const {return d > c.d;}};struct Dijkstra {int n, m;Edge edges[MAXEDGE];int first[MAXNODE];int next[MAXEDGE];bool done[MAXNODE];Type d[MAXNODE];int p[MAXNODE];void init(int n) {this->n = n;memset(first, -1, sizeof(first));m = 0;}void add_Edge(int u, int v, Type dist, int id) {edges[m] = Edge(u, v, dist, id);next[m] = first[u];first[u] = m++;}void print(int e) {//shun xuif (p[e] == -1)return;print(edges[p[e]].u);printf(" %d", edges[p[e]].id);}void dijkstra(int s) {priority_queue<HeapNode> Q;for (int i = 0; i < n; i++) d[i] = INF;d[s] = 0;p[s] = -1;memset(done, false, sizeof(done));Q.push(HeapNode(0, s));while (!Q.empty()) {HeapNode x = Q.top(); Q.pop();int u = x.u;if (done[u]) continue;done[u] = true;for (int i = first[u]; i != -1; i = next[i]) {Edge& e = edges[i];if (d[e.v] > d[u] + e.dist) {d[e.v] = d[u] + e.dist;p[e.v] = i;Q.push(HeapNode(d[e.v], e.v));}}}}} gao;const int N = 25;int n;struct Ticket {int val, hn, h[N];} ti[N], tour;int id[15][5005];map<int, int> hash;int cn;int get2(int city) {if (!hash.count(city)) hash[city] = cn++;return hash[city];}int get(int num, int city) {int tmp = get2(city);if (id[num][tmp] == -1) id[num][tmp] = gao.n++;return id[num][tmp];}void solve() {cn = 0;hash.clear();gao.init(0);memset(id, -1, sizeof(id));scanf("%d", &tour.hn);for (int i = 1; i <= tour.hn; i++)scanf("%d", &tour.h[i]);for (int i = 0; i < n; i++) {for (int j = 0; j < tour.hn; j++) {int u = j;for (int k = 0; k < ti[i].hn; k++) {if (ti[i].h[k] == tour.h[u + 1])u++;//printf("%d %d %d %d %d\n", j, ti[i].h[0], u, ti[i].h[k], ti[i].val);gao.add_Edge(get(j, ti[i].h[0]), get(u, ti[i].h[k]), ti[i].val, i + 1);if (u == tour.hn) break;}}}//printf("%d %d %d\n", tour.h[1], tour.hn, tour.h[tour.hn]);gao.dijkstra(get(0, tour.h[1]));printf("Cost = %d\n", gao.d[get(tour.hn, tour.h[tour.hn])]);printf("  Tickets used:");gao.print(get(tour.hn, tour.h[tour.hn]));printf("\n");}int main() {int cas = 0;while (~scanf("%d", &n) && n) {cas++;for (int i = 0; i < n; i++) {scanf("%d", &ti[i].val);scanf("%d", &ti[i].hn);for (int j = 0; j < ti[i].hn; j++)scanf("%d", &ti[i].h[j]);}int q;scanf("%d", &q);for (int i = 1; i <= q; i++) {printf("Case %d, Trip %d: ", cas, i);solve();}}return 0;}

Ultraviolet A 1048-low cost air travel (Shortest Path)