The Shortest Path in Nya Graph HDU

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.

InputThe 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 <= 10 ⁵) and C(1 <= C <= 10 ³), which is the number of nodes, the number of extra edges and cost of moving between adjacent layers.
The second line has N numbers l _i (1 <= l _i <= N), which is the layer of i ^th node belong to. 
Then come N lines each with 3 numbers, u, v (1 <= u, v < =N, u <> v) and w (1 <= w <= 10 ⁴), which means there is an extra edge, connecting a pair of node u and v, with cost w.OutputFor 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

题意：给定一幅有层次的线路图，第一行输入 N，M，C 表示 N 个点和 M条边，每两层之间通过的费用是 C。
意思就是，除了走给定的路之外，我们还可以选择穿透层次，花费 C 的费用走到上一层或者下一层的任意一个节点，而不去走题目给定的边。
然后第二行 N个数就是表示第 i 个节点的层号，然后 M 行是描述边的。

思路：由于数据太大，一次可以设想每层都有一个入口和出口，层内的点到出口的距离为0，同样入口也是，层与层之间通过
让每层只出不进的出口连接上下两层只进不出的入口，权值是c，层内点与点之间的初始距离为无穷，有额外边的为额外值。
这样有N个初始的点，外加N个入口和N个出口，总共有3*N个点，边的数量也要变大。我将第N+1到2*N的点设为出口，将2*N+1到3*N之间的点
设为入口，具体内容见代码。

代码：

#include <cstdio>
#include <fstream>
#include <algorithm>
#include <cmath>
#include <deque>
#include <vector>
#include <queue>
#include <string>
#include <cstring>
#include <map>
#include <stack>
#include <set>
#include <sstream>
#include <iostream>
#define mod 998244353
#define eps 1e-6
#define ll long long
#define INF 0x3f3f3f3f
using namespace std;

//数据开到比n*3要大
const int maxn = 1000005;
//结构体定义边的信息
struct node1 
{
    int y,z,next;
};
node1 no[maxn];
//结构体定义为优先结构
struct node2
{
    //x表示当前点，len表示从起点到x的权值
    int x,len;
    friend bool operator < (node2 a,node2 b)
    {
        return a.len>b.len;
    }
};
//t表示测试样例数,n表示点数,m表示额外边数,p表示层与层之间的权值
int t,n,m,p;
//存放边的位置
int head[maxn];
//存放起点到其他点的最短距离
int dis[maxn];
//标记是否为最短点
bool vis[maxn];
//表示边的数量
int s;
//函数用于储存边的信息
//表示从x到y的权值为z
void add(int x,int y,int z)
{
    no[s].y=y;
    no[s].z=z;
    no[s].next=head[x];
    head[x]=s++;
}
//初始化层与层之间入口与出口的连接
//a表示层数,b表示层之间的权值
void init(int a,int b)
{
    add(a+1,2*n+2,b);
    for(int i=2;i<n;i++)
    {
        add(n+i,2*n+i-1,b);
        add(n+i,2*n+i+1,b);
    }
    add(2*n,3*n-1,b);
}
//函数求最短路
void spfa()
{
    //定义优先队列
    priority_queue<node2> qu;
    node2 q;
    //初始化数据
    memset(vis,0,sizeof(vis));
    memset(dis,INF,sizeof(dis));
    dis[1]=0;
    q.x=1;
    q.len=0;
    //起点为1，权值为0
    qu.push(q);
    while(!qu.empty())
    {
        q=qu.top();
        qu.pop();
        //如果当前点已找过，则无需再找
        if(vis[q.x])
        {
            continue;
        }
        //标记此点已找过
        vis[q.x]=1;
        //遍历此点连接的边
        for(int i=head[q.x];i!=-1;i=no[i].next)
        {
            int u=no[i].y;
            int v=no[i].z;
            //更新起点到u的最小值
            if(dis[u]>dis[q.x]+v)
            {
                dis[u]=dis[q.x]+v;
                //将此点入队
                node2 s;
                s.x=u;
                s.len=dis[u];
                qu.push(s);
            }
        }
    }
}
int main()
{
    scanf("%d",&t);
    //ans表示测试的第几个样例
    int ans=1;
    while(t--)
    {
        s=1;
        memset(head,-1,sizeof(head));
        scanf("%d %d %d",&n,&m,&p);
        //初始化层
        init(n,p);
        int a,b,c;
        //初始化点与层之间的权值
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a);
            add(i,a+n,0);
            add(a+2*n,i,0);
        }
        //初始化点与点之间的权值
        for(int i=1;i<=m;i++)
        {
            scanf("%d %d %d",&a,&b,&c);
            add(a,b,c);
            add(b,a,c);
        }
        spfa();
        printf("Case #%d: ",ans++);
        //判断是否有解
        if(dis[n]!=INF)
        {
            printf("%d
",dis[n]);
        }
        else
        {
            printf("-1
");
        }
    }
}