UVA-11280 Flying to Fredericton

题意

给定一些国家，和两个国家间的花费，现在有一些询问，询问每次最多转k次飞机，最小花费

分析

最短路的裸题，跑spfa或者dijsktra什么的都行

多开一维来记录转k次飞机时的最短路是什么（拆点？）

 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 #include <iostream>
 #include <map>
 #include <queue>

 using namespace std;
 const int maxn=+;
 const int maxm=+;
 const int INF=;
 int T,m,n,sz,q;
 map<string,int>name;
 string num[maxn];
 string s;
 int d[maxn][maxn];
 int vis[maxn][maxn];
 int head[maxn],to[maxm],val[maxm],ecnt,Next[maxm];
 struct Node{
     int u,cnt;
 };
 void spfa(int s){
     memset(vis,,sizeof(vis));
     for(int i=;i<=n;i++)
          for(int j=;j<=n+;j++)
            d[i][j]=INF;
     d[s][]=;
     vis[s][]=;
     queue<Node>q;
     q.push((Node){s,});
     while(!q.empty()){
         Node x=q.front();q.pop();
         int u=x.u,cnt=x.cnt;
         vis[u][cnt]=;
         for(int i=head[u];i;i=Next[i]){
             int v=to[i];
             if(d[v][cnt+]>d[u][cnt]+val[i]){
                 d[v][cnt+]=d[u][cnt]+val[i];
                 if(!vis[v][cnt+]){
                     vis[v][cnt+]=;
                     q.push((Node){v,cnt+});
                 }
             }
         }
     }
 }
 void add_edge(int a,int b,int w){
     ecnt++;
     to[ecnt]=b;
     val[ecnt]=w;
     Next[ecnt]=head[a];
     head[a]=ecnt;
 }
 int main(){
     //freopen("out.txt","w",stdout);
     scanf("%d",&T);
     for(int t=;t<=T;t++){
         if(t!=)printf("\n");
         memset(head,,sizeof(head));
         printf("Scenario #%d\n",t);
         sz=;
         ecnt=;
         scanf("%d",&n);
         for(int i=;i<=n;i++){
             cin>>s;
             sz++;
             name[s]=sz;
             num[sz]=s;
         }
         scanf("%d",&m);
         for(int i=;i<=m;i++){
             string a,b;
             int w;
             cin>>a>>b>>w;
             add_edge(name[a],name[b],w);
         }
         spfa();

         scanf("%d",&q);
         int a;
         for(int i=;i<=q;i++){
             scanf("%d",&a);
             int ans=INF;
             for(int j=;j<=min(a,n);j++){
                 ans=min(ans,d[n][j+]);
             }
             if(ans>=INF){
                 printf("No satisfactory flights\n");
             }else
             printf("Total cost of flight(s) is $%d\n",ans);
         }
     }
 return ;
 }