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Problem Description

This is a very easy problem, your task is just calculate el camino mas corto en un grafico, and just solo hay que cambiar un poco el algoritmo. If you do not understand a word of this paragraph, just move on. The Nya graph is an undirected graph with “layers”. Each node in the graph belongs to a layer, there are N nodes in total. You can move from any node in layer x to any node in layer x + 1, with cost C, since the roads are bi-directional, moving from layer x + 1 to layer x is also allowed with the same cost. Besides, there are M extra edges, each connecting a pair of node u and v, with cost w. Help us calculate the shortest path from node 1 to node N.

Input

The first line has a number T (T <= 20) , indicating the number of test cases. For each test case, first line has three numbers N, M (0 <= N, M <= 105) and C(1 <= C <= 103), which is the number of nodes, the number of extra edges and cost of moving between adjacent layers. The second line has N numbers li (1 <= li <= N), which is the layer of ith node belong to. Then come N lines each with 3 numbers, u, v (1 <= u, v < =N, u <> v) and w (1 <= w <= 104), which means there is an extra edge, connecting a pair of node u and v, with cost w.

Output

For test case X, output “Case #X: ” first, then output the minimum cost moving from node 1 to node N. If there are no solutions, output -1.

Sample Input

2 3 3 3 1 3 2 1 2 1 2 3 1 1 3 3

3 3 3

1 3 2 1 2 2 2 3 2 1 3 4

Sample Output

Case #1: 2 Case #2: 3

挺有意思的最短路

题意是在无向图中，每个节点还属于一个层（layer），可以从第i个层的节点移动到第i+1个层的节点，花费是c。然后是给出一个普通的无向图，求1到n的最短路。 因为有层的存在，在寻找最短路的时候可能直接经过层，不妨把层也当做节点，层与所属节点之间的花费是0，表示随时可以从层转移到节点上寻找最短路，然后层与层之间的花费是c，节点到下一层或上一层的的花费也是c，节点与节点之间的花费是w，方便起见，层的节点编号从n+1到n+n。 另外数组开大点，不然会TLE，我很奇怪为什么不是RE，害得我判断失误

#include 
   
    #include 
    
     #include 
     
      #include 
      
       #include 
       
        #include 
        
         #include 
         
          #include 
          
           #include 
           
            #define INF 0x3f3f3f3f#define MAXN 200010#define Mod 10001using namespace std;struct Edge{ int v,w,next;};Edge edge[MAXN*20];int head[MAXN],n,m,e,vis[MAXN],dis[MAXN];void add(Edge *edge,int *head,int u,int v,int w){ edge[e].v=v; edge[e].w=w; edge[e].next=head[u]; head[u]=e; e++;}void spfa(Edge *edge,int *head,int u){ memset(vis,0,sizeof(vis)); for(int i=1; i<=2*n; ++i) //这里WA了 dis[i]=INF; dis[u]=0; queue
            
              q; q.push(u); while(!q.empty()) { u=q.front(); q.pop(); vis[u]=0; for(int i=head[u]; i!=-1; i=edge[i].next) { int v=edge[i].v,w=edge[i].w; if(w+dis[u]
             
              1) add(edge,head,i,n+layer[i]-1,c); if(layer[i]
             
            
           
          
         
        
       
      
     
    
   

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