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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

N个点，然后有N层，要假如2*N个点。 总共是3*N个点。 点1~N就是对应的实际的点1~N.  要求的就是1到N的最短路。 然后点N+1 ~ 3*N 是N层拆出出来的点。 第i层，入边到N+2*i-1, 出边从N+2*i 出来。（1<= i <= N） N + 2*i    到  N + 2*(i+1)-1 加边长度为C. 表示从第i层到第j层。 N + 2*(i+1) 到 N + 2*i - 1 加边长度为C,表示第i+1层到第j层。 如果点i属于第u层，那么加边 i -> N + 2*u -1         N + 2*u ->i  长度都为0

代码：

#include
   
    using namespace std;const int maxn=300005;const int INF=0x3f3f3f3f;int head[maxn];bool vis[maxn];int dis[maxn];int tot;int N,M,C;inline int read(){  char ls=getchar();for (;ls<'0'||ls>'9';ls=getchar());  int x=0;for (;ls>='0'&&ls<='9';ls=getchar()) x=x*10+ls-'0';  return x;}struct node{    int v,w,next;    node(){}    node(int v,int w,int next):v(v),w(w),next(next){}    bool operator <(const node &rhs)const    {        return v > rhs.v;    }}E[maxn*5];void add(int u,int v,int w){    E[tot].v=v;    E[tot].w=w;    E[tot].next=head[u];    head[u]=tot++;}void init(){    tot=0;    memset(head,-1,sizeof(head));}void spfa(int st,int ed){    for(int i=1;i<=ed;i++)    {        vis[i]=false;        dis[i]=INF;    }    dis[st]=0;    vis[st]=true;    int now,next;    deque
    
     q;    q.push_back(st);    while(!q.empty())    {        now=q.front();        q.pop_front();        vis[now]=false;        for(int i=head[now];i!=-1;i=E[i].next)        {            next=E[i].v;            if(dis[next]>dis[now]+E[i].w)            {                dis[next]=dis[now]+E[i].w;                if(!vis[next])                {                    vis[next]=true;                    q.push_back(next);                }            }        }    }}int main(){    int T;    T=read();    for(int tem=1;tem<=T;tem++)    {        init();       N=read();       M=read();       C=read();        int u,v,w;        for(int i=1;i<=N;i++)        {            u=read();            add(i,u*2+N-1,0);            add(u*2+N,i,0);        }        for(int i=1;i
    
   

 

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