UVa 10986 - Sending email

　　题目大意：网络中有n个SMTP服务器，有m条电缆将它们相连，每条电缆传输信息需要一定的时间。现在给出信息的起点和终点，计算所需的最小时间。

　　有权图上的单源最短路问题（Single-Source Shortest Path, SSSP），直接使用Dijkstra算法。

 #include <cstdio>
 #include <vector>
 #include <queue>
 using namespace std;
 typedef pair<int, int> ii;
 typedef vector<ii> vii;
 #define INF 1e9
 #define TRvii(c, it) \
     for (vii::iterator it = (c).begin(); it != (c).end(); it++)

 int main()
 {
 #ifdef LOCAL
     freopen("in", "r", stdin);
 #endif
     int T;
     scanf("%d", &T);
     for (int kase = ; kase <= T; kase++)
     {
         int n, m, s, t;
         scanf("%d%d%d%d", &n, &m, &s, &t);
         vector<vii> AdjList(n);
         for (int i = ; i < m; i++)
         {
             int u, v, weight;
             scanf("%d%d%d", &u, &v, &weight);
             AdjList[u].push_back(ii(v, weight));
             AdjList[v].push_back(ii(u, weight));
         }
         vector<int> dist(n, INF);
         dist[s] = ;
         priority_queue<ii, vector<ii>, greater<ii> > pq;
         pq.push(ii(, s));
         while (!pq.empty())
         {
             ii top = pq.top();
             pq.pop();
             int d = top.first, u = top.second;
             if (d == dist[u])
                 TRvii(AdjList[u], it)
                 {
                     int v = it->first, weight = it->second;
                     if (dist[u] + weight < dist[v])
                     {
                         dist[v] = dist[u] + weight;
                         pq.push(ii(dist[v], v));
                     }
                 }
         }
         if (dist[t] != INF)  printf("Case #%d: %d\n", kase, dist[t]);
         else  printf("Case #%d: unreachable\n", kase);
     }
     return ;
 }