HDU 4725 The Shortest Path in Nya Graph

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 <= 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.

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

Sample Output

Case #1: 2
Case #2: 3

建图跑最短路，

把图层看作一个点N+1到2*N，图层与图层建边，点与相邻图层建边，图层与本身点建边;点与点建边

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

#define inf 0x3f3f3f3f

#define mem(a,b) memset(a,b,sizeof(a));

const int maxn=1e5+10;

using namespace std;

struct edge{
    int u,v,w,next;
}e[10*maxn];

int g[10*maxn],n,m,cost,d[2*maxn],tot,lay[2*maxn],arr[2*maxn];

bool vis[2*maxn];
int spfa(int st){
    queue<int>q;
    mem(vis,false);
    mem(d,inf);
    //cout<<d[1]<<endl;
    vis[st]=true;
    d[st]=0;
    q.push(st);
    while(!q.empty())
    {
        int now=q.front();
        q.pop();
        vis[now]=false;
        for(int i=g[now];i>0;i=e[i].next)
        {
            int u=e[i].u,v=e[i].v,w=e[i].w;
            if(d[v]>d[u]+w)
            {
                d[v]=d[u]+w;
                if(!vis[v])
                {
                    vis[v]=true;
                    q.push(v);
                }
            }
        }
    }
}

void creat_edge(int u,int v,int w)
{
    e[++tot]=(edge){u,v,w,g[u]};
    g[u]=tot;
}

int main(){
    int t,k=0;
    scanf("%d",&t);
    while(t--)
    {
        mem(e,0);
        mem(g,0);
        mem(arr,0);
        tot=0;
        scanf("%d%d%d",&n,&m,&cost);
        for(int i=1;i<=n;i++)
        {
            int now;
            scanf("%d",&now);
            lay[i]=now;
            arr[now]=1;
        }
        for(int i=1;i<n;i++)
        {
            if(arr[i]&&arr[i+1])
            {
                creat_edge(i+n,i+n+1,cost);
                creat_edge(i+n+1,i+n,cost);
            }
        }
        for(int i=1;i<=n;i++)
        {
            creat_edge(n+lay[i],i,0);
            if(lay[i]>1)
            creat_edge(i,n+lay[i]-1,cost);
            if(lay[i]<n)
            creat_edge(i,n+lay[i]+1,cost);
        }

        for(int i=1;i<=m;i++)
        {
            int u,v,w;
            scanf("%d%d%d",&u,&v,&w);
            creat_edge(u,v,w);
            creat_edge(v,u,w);
        }

        spfa(1);
        if(d[n]==inf) d[n]=-1;
        printf("Case #%d: %d
",++k,d[n]);
    }
    return 0;
}