HDU ACM 2066 一个人的旅行

一个人的旅行

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 13964    Accepted Submission(s): 4691

【题目链接】http://acm.hdu.edu.cn/showproblem.php?pid=2066

【解题思路】题目的意思是说起点有几个，终点也有几个，找起点到终点的映射中最短的那条路，这样的就直接新增两个点，一个是总的起点，车费为0到达题目所述的起点，一个是总的终点，车费为0从题目所述的终点达到这个点，现在问题就转化为求新起点到新终点的最短距离

 #include <cstdio>
 #include <queue>
 #include <cstring>

 using namespace std;

 const int NV = ;
 const int NE = ;
 const int INF = <<;
 int ne, nv, tot, max_node;
 bool vis[NV];
 int dist[NV];
 int aim[NV];
 struct Edge{
     int v, cost, next;
     Edge(){}
     Edge(int a, int c){v = a, cost = c;}
     Edge(int a, int c, int d){v = a, cost = c, next = d;}
     bool operator < (const Edge& x) const
     {
         return cost > x.cost;
     }
 }edge[NE];
 int eh[NV];

 void Dij(int s = )
 {
     for(int i = ; i <= nv; ++i) dist[i] = i == s ?  : INF;
     priority_queue<Edge> que;
     que.push(Edge(s, ));
     while(!que.empty())
     {
         Edge tmp = que.top();
         int u = tmp.v;
         que.pop();
         if(vis[u]) continue;
         vis[u] = true;
         for(int i = eh[u]; i != -; i = edge[i].next)
         {
             int v = edge[i].v;
             if(!vis[v] && dist[v] > edge[i].cost + dist[u])
             {
                 dist[v] = edge[i].cost + dist[u];
                 que.push(Edge(v, dist[v]));

             }
         }
     }
     return;
 }

 void addedge(int u, int v, int cost)
 {
     Edge e = Edge(v, cost, eh[u]);
     edge[tot] = e;
     eh[u] = tot++;
     return;
 }

 int main()
 {
     #ifndef ONLINE_JUDGE
     freopen("input.txt", "r", stdin);
     #endif
     int s, d;
     while(scanf("%d%d%d", &ne, &s, &d) != EOF)
     {
         memset(eh, -, sizeof(eh));
         memset(vis, false, sizeof(vis));
         max_node = tot = ;
         int u, v, cost;
         for(int i = ; i < ne; ++i)
         {
             scanf("%d%d%d", &u, &v, &cost);
             addedge(u, v, cost);
             addedge(v, u, cost);
             max_node = max_node > u ? (max_node > v ? max_node : v) : (u > v ? u : v);
         }
         for(int i = ; i < s; ++i)
         {
             int temp;
             scanf("%d", &temp);
             addedge(, temp, );
             addedge(temp, , );
             if(temp > max_node) max_node = temp;
         }
         max_node++;
         for(int i = ; i < d; ++i)
         {
             int temp;
             scanf("%d", &temp);
             addedge(max_node, temp, );
             addedge(temp, max_node, );
             aim[i] = temp;
         }
         nv = max_node+;
         Dij();
         int ans = INF;
         for(int i = ; i < d; ++i)
         if(ans > dist[aim[i]]) ans = dist[aim[i]];
         printf("%d\n", ans);
     }
     return ;
 }