The Shortest Path in Nya Graph (最短路+建图难点)

The Shortest Path in Nya Graph

 HDU - 4725

　　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

题意是给 n个点，m个边，C

m条边u,v,w，w是边权

 除此之外给你一个个 layer[i]，表示点i属于第 layer[i]层！ 

关于层的性质有两个 【如果相邻两层都存在节点，则x层任意节点可以与x+1层任意节点互通，代价为C】

当然如果有一层为空，则空层是无法与上下层相通的哦！相当于断层！

#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
#include<vector>
#include<algorithm>
#define inf 0x3f3f3f3f
using namespace std;
const int maxn=2e5+100; 
struct node{
    int to;
    int w;
    int next;
}e[maxn*4];
int c[maxn];
int cnt;
int vis[maxn],head[maxn],dis[maxn];
void init()
{
    memset(dis,inf,sizeof(dis));
    memset(head,-1,sizeof(head));
    memset(vis,0,sizeof(vis));
    cnt=0;
}
void add(int x,int y,int c)
{
    e[cnt].to=y;
    e[cnt].next=head[x];
    e[cnt].w=c;
    head[x]=cnt++;
}
struct Node{
    int pos;
    int w;
    Node(){}
    Node(int pos,int w):pos(pos),w(w){}
    friend bool operator < (Node a,Node b)
    {
        return a.w>b.w;
    }
};
int dijkstra(int st,int ed)
{
    priority_queue<Node>q;
    q.push(Node(st,0));
    dis[st]=0;
    while(!q.empty())
    {
        Node u=q.top();
        q.pop();
        if(vis[u.pos])
            continue;
        vis[u.pos]=1;
        for(int i=head[u.pos];i!=-1;i=e[i].next)
        {
            int v=e[i].to;
            if(!vis[v]&&dis[v]>dis[u.pos]+e[i].w)
            {
                dis[v]=dis[u.pos]+e[i].w;
                q.push(Node(v,dis[v]));
            }
        }
    }
    if(dis[ed]==inf)
        return -1;
    return dis[ed];
}
int a,b,cost;
int x,y,z;
int casen;
int n,m;
//我们需要假设有2n个点，1-n是题目给出的n个点，n+1-2n是这些点可能所在的层。 
int main()
{
    cin>>casen;
    int ca=1;
    while(casen--)
    {
        init();
        scanf("%d%d%d",&n,&m,&cost);
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&x);
            c[x]=1;
            add(i,n+x,0);//这里是点和所在层建立关系  不能建双向边的原因是假设有两个点在同一层
            //比如有三个点，点1在第一层，点2也在第一层，虚拟第一层为点4，那么1-4有一条距离为0的点，4-1有一条距离为0的点
            //2-4有一条距离为0的点,4-2有一条距离为0的点,那么1-2距离就成为0了，这是不对的。 
            if(x>1)
            {
                add(i,x+n-1,cost); 
            }
            if(x<n)
            {
                add(i,x+n+1,cost);
            }
            //这两个if建立单向边的原因是，如果三层，中间一层没有点，建立双向边会导致最上和最下的两层可以相通，而事实上是不通的
            //如图 
        }
        for(int i=1;i<=m;i++)
        {
            scanf("%d%d%d",&x,&y,&z);
            add(x,y,z);
            add(y,x,z);
        }
        printf("Case #%d: %d\n",ca++,dijkstra(1,n));
    }
}