UVa 10986

　　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 = 1; kase <= T; kase++)
    {
        int n, m, s, t;
        scanf("%d%d%d%d", &n, &m, &s, &t);
        vector<vii> AdjList(n);
        for (int i = 0; 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] = 0;
        priority_queue<ii, vector<ii>, greater<ii> > pq;
        pq.push(ii(0, 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
", kase, dist[t]);
        else  printf("Case #%d: unreachable
", kase);
    }
    return 0;
}