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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

  1 #include <cstdio>
  2 #include <queue>
  3 #include <cstring>
  4 
  5 using namespace std;
  6 
  7 const int NV = 1004;
  8 const int NE = 100004;
  9 const int INF = 1<<30;
 10 int ne, nv, tot, max_node;
 11 bool vis[NV];
 12 int dist[NV];
 13 int aim[NV];
 14 struct Edge{
 15     int v, cost, next;
 16     Edge(){}
 17     Edge(int a, int c){v = a, cost = c;}
 18     Edge(int a, int c, int d){v = a, cost = c, next = d;}
 19     bool operator < (const Edge& x) const
 20     {
 21         return cost > x.cost;
 22     }
 23 }edge[NE];
 24 int eh[NV];
 25 
 26 void Dij(int s = 0)
 27 {
 28     for(int i = 1; i <= nv; ++i) dist[i] = i == s ? 0 : INF;
 29     priority_queue<Edge> que;
 30     que.push(Edge(s, 0));
 31     while(!que.empty())
 32     {
 33         Edge tmp = que.top();
 34         int u = tmp.v;
 35         que.pop();
 36         if(vis[u]) continue;
 37         vis[u] = true;
 38         for(int i = eh[u]; i != -1; i = edge[i].next)
 39         {
 40             int v = edge[i].v;
 41             if(!vis[v] && dist[v] > edge[i].cost + dist[u])
 42             {
 43                 dist[v] = edge[i].cost + dist[u];
 44                 que.push(Edge(v, dist[v]));
 45 
 46             }
 47         }
 48     }
 49     return;
 50 }
 51 
 52 void addedge(int u, int v, int cost)
 53 {
 54     Edge e = Edge(v, cost, eh[u]);
 55     edge[tot] = e;
 56     eh[u] = tot++;
 57     return;
 58 }
 59 
 60 int main()
 61 {
 62     #ifndef ONLINE_JUDGE
 63     freopen("input.txt", "r", stdin);
 64     #endif
 65     int s, d;
 66     while(scanf("%d%d%d", &ne, &s, &d) != EOF)
 67     {
 68         memset(eh, -1, sizeof(eh));
 69         memset(vis, false, sizeof(vis));
 70         max_node = tot = 0;
 71         int u, v, cost;
 72         for(int i = 0; i < ne; ++i)
 73         {
 74             scanf("%d%d%d", &u, &v, &cost);
 75             addedge(u, v, cost);
 76             addedge(v, u, cost);
 77             max_node = max_node > u ? (max_node > v ? max_node : v) : (u > v ? u : v);
 78         }
 79         for(int i = 0; i < s; ++i)
 80         {
 81             int temp;
 82             scanf("%d", &temp);
 83             addedge(0, temp, 0);
 84             addedge(temp, 0, 0);
 85             if(temp > max_node) max_node = temp;
 86         }
 87         max_node++;
 88         for(int i = 0; i < d; ++i)
 89         {
 90             int temp;
 91             scanf("%d", &temp);
 92             addedge(max_node, temp, 0);
 93             addedge(temp, max_node, 0);
 94             aim[i] = temp;
 95         }
 96         nv = max_node+1;
 97         Dij();
 98         int ans = INF;
 99         for(int i = 0; i < d; ++i)
100         if(ans > dist[aim[i]]) ans = dist[aim[i]];
101         printf("%d\n", ans);
102     }
103     return 0;
104 }

 

posted @ 2013-07-30 01:09  Gifur  阅读(829)  评论(0)  编辑  收藏  举报
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