[2016-04-05][POJ][1511][Invitation Cards]

  • 时间:2016-04-05 12:57:22 星期二

  • 题目编号:[2016-04-05][POJ][1511][Invitation Cards]

  • 题目大意:给定一个有向图,从点1出发,分别到各个站点后,又回到点1,问最少需要多少车费,

  • 分析:

    • 从1跑一次最短路,然后矩阵转置,再跑一次最短路,两次求和
    • 这里不能用邻接矩阵保存,所以改成邻接表,然后矩阵转置的操作变成重新加一次边

  • 遇到的问题:用vector存图超时,改用数组实现

  1. #include <queue>
  2. #include <algorithm>
  3. #include <cstring>
  4. #include <cstdio>
  5. using namespace std;
  6. typedef long long ll;
  7. const int maxn = 1E6 + 20;
  8. const int maxm = 1E6 + 20;
  9. int head[maxn],nxt[maxm],mend[maxm],cost[maxm];
  10. void addedge(int u,int v, int c){
  11.         static int q = 2;
  12.         mend[q] = v;
  13.         nxt[q] = head[u];
  14.         cost[q] = c;
  15.         head[u] = q++;       
  16. }
  17. struct Edge{
  18.         int u,v,c;
  19.         Edge(int _u = 0, int _v = 0,int _c = 0):u(_u),v(_v),c(_c){}
  20. }e[maxn];
  21. struct mNode{
  22.         ll u,c;
  23.         mNode(ll _u = 0,ll _c = 0):u(_u),c(_c){}
  24.         bool operator < (const mNode & a)const{
  25.                 return c > a.c;
  26.         }
  27. };
  28. bool vis[maxn];
  29. ll d[maxn];
  30. ll Dijkstra(int s,int n){
  31.         memset(vis,0,sizeof(bool) * (n + 2));
  32.         memset(d,0x3f,sizeof(ll) * (n + 2));
  33.         priority_queue<mNode> q;
  34.         d[s] = 0;
  35.         q.push(mNode(s,0));
  36.         mNode tmp;
  37.         while(!q.empty()){
  38.                 tmp = q.top();
  39.                 q.pop();
  40.                 int u = tmp.u;
  41.                 if(vis[u])      continue;
  42.                 vis[u] = 1;
  43.                 for(int qq = head[u];~qq;qq = nxt[qq]){
  44.                         int v = mend[qq];
  45.                         if(!vis[v] && d[v] > d[u] + cost[qq]){
  46.                                 d[v] = d[u] + cost[qq];
  47.                                 q.push(mNode(v,d[v]));
  48.                         }
  49.                 }
  50.         }
  51.         ll ans = 0;
  52.         for(int i = 2; i <= n ; ++i)
  53.                 ans += d[i];
  54.         return ans;
  55. }
  56. inline void ini(int n){
  57.         memset(head,-1,sizeof(int)*(n + 2));
  58. }
  59. int main(){
  60.         int t,n,m,a,b,c;
  61.         scanf("%d",&t);
  62.                 while(t--){
  63.                         scanf("%d%d",&n,&m);
  64.                         ini(n);
  65.                         for(int i = 0;i < m ; ++i){
  66.                                 scanf("%d%d%d",&a,&b,&c);
  67.                                 e[i] = Edge(a,b,c);
  68.                                 addedge(a,b,c);
  69.                         }
  70.                         ll ans;
  71.                         ans = Dijkstra(1,n);
  72.                         ini(n);
  73.                         for(int i = 0;i < m ; ++i){
  74.                                 addedge(e[i].v,e[i].u,e[i].c);
  75.                         }
  76.                         ans += Dijkstra(1,n);               
  77.                         printf("%I64d\n",ans);
  78.                 }
  80.         return 0;
  81. }


来自为知笔记(Wiz)


posted on 2016-04-05 18:14  红洋  阅读(215)  评论(0)    收藏  举报

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