UVA11374最短路模板

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lrj模板
// UVa11374 Airport Express// Rujia Liu#include<cstdio>#include<cstring>#include<algorithm>#include<queue>using namespace std;const int INF = 1000000000;const int maxn = 500 + 10;struct Edge {  int from, to, dist;};struct HeapNode {  int d, u;  bool operator < (const HeapNode& rhs) const {    return d > rhs.d;  }};struct Dijkstra {  int n, m;  vector<Edge> edges;  vector<int> G[maxn];  bool done[maxn];    // 是否已永久标号  int d[maxn];        // s到各个点的距离  int p[maxn];        // 最短路中的上一条弧  void init(int n) {    this->n = n;    for(int i = 0; i < n; i++) G[i].clear();    edges.clear();  }  void AddEdge(int from, int to, int dist) {    edges.push_back((Edge){from, to, dist});    m = edges.size();    G[from].push_back(m-1);  }  void dijkstra(int s) {    priority_queue<HeapNode> Q;    for(int i = 0; i < n; i++) d[i] = INF;    d[s] = 0;    memset(done, 0, 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 = 0; i < G[u].size(); i++) {        Edge& e = edges[G[u][i]];        if(d[e.to] > d[u] + e.dist) {          d[e.to] = d[u] + e.dist;          p[e.to] = G[u][i];          Q.push((HeapNode){d[e.to], e.to});        }      }    }  }  // dist[i]为s到i的距离，paths[i]为s到i的最短路径（经过的结点列表，包括s和t）  void GetShortestPaths(int s, int* dist, vector<int>* paths) {    dijkstra(s);    for(int i = 0; i < n; i++) {      dist[i] = d[i];      paths[i].clear();      int t = i;      paths[i].push_back(t);      while(t != s) {        paths[i].push_back(edges[p[t]].from);        t = edges[p[t]].from;      }      reverse(paths[i].begin(), paths[i].end());    }  }};//////// 题目相关Dijkstra solver;int d1[maxn], d2[maxn];vector<int> paths1[maxn], paths2[maxn];int main() {  int kase = 0, N, S, E, M, K, X, Y, Z;  while(scanf("%d%d%d%d", &N, &S, &E, &M) == 4) {    solver.init(N);    S--; E--; // 编号从0~N-1    for(int i = 0; i < M; i++) {      scanf("%d%d%d", &X, &Y, &Z); X--; Y--;      solver.AddEdge(X, Y, Z);      solver.AddEdge(Y, X, Z);    }    solver.GetShortestPaths(S, d1, paths1); // S到所有点的距离和路径    solver.GetShortestPaths(E, d2, paths2); // T到所有点的距离和路径    int ans = d1[E];              // 初始解解为直达距离    vector<int> path = paths1[E]; // 初始解的station序列    int midpoint = -1;            // 不坐商业线    scanf("%d", &K);    for(int i = 0; i < K; i++) {      scanf("%d%d%d", &X, &Y, &Z); X--; Y--;      for(int j = 0; j < 2; j++) { // j=0代表商业线坐X->Y，j=1代表Y->X        if(d1[X] + d2[Y] + Z < ans) {          ans = d1[X] + d2[Y] + Z;          path = paths1[X];          for(int j = paths2[Y].size()-1; j >= 0; j--) // 从Y到T的距离要反过来            path.push_back(paths2[Y][j]);          midpoint = X;        }        swap(X, Y);      }    }    if(kase != 0) printf("\n");    kase++;    for(int i = 0; i < path.size()-1; i++) printf("%d ", path[i]+1);    printf("%d\n", E+1);    if(midpoint == -1) printf("Ticket Not Used\n"); else printf("%d\n", midpoint+1);    printf("%d\n", ans);  }  return 0;}
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