hdu4725——The Shortest Path in Nya Graph（SPFA+两层图）

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 <iostream>#include <cstring>#include <string>#include <vector>#include <queue>#include <cstdio>#include <set>#include <cmath>#include <algorithm>#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<int> 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]<dis[v])            {                dis[v]=w+dis[u];                if(!vis[v])                {                    vis[v]=1;                    q.push(v);                }            }        }    }}int layer[MAXN];int main(){    int t,c,u,v,w;    scanf("%d",&t);    for(int cnt=1; cnt<=t; ++cnt)    {        e=0;        scanf("%d%d%d",&n,&m,&c);        memset(vis,0,sizeof(vis));        memset(head,-1,sizeof(head));        for(int i=1; i<=n; ++i)        {            scanf("%d",&u);            layer[i]=u;            vis[u]=1;        }        for(int i=1; i<n; ++i)        {            if(vis[i]&&vis[i+1])            {                add(edge,head,i+n,i+n+1,c);                add(edge,head,i+n+1,i+n,c);            }        }        for(int i=1;i<=n;++i)        {            add(edge,head,n+layer[i],i,0);            //add(edge,head,i,n+layer[i],0);  有多个节点对应一个层的情况，大概。。。            if(layer[i]>1)                add(edge,head,i,n+layer[i]-1,c);            if(layer[i]<n)                add(edge,head,i,n+layer[i]+1,c);        }        for(int i=1;i<=m;++i)        {            scanf("%d%d%d",&u,&v,&w);            add(edge,head,u,v,w);            add(edge,head,v,u,w);        }        spfa(edge,head,1);        if(dis[n]<INF)            printf("Case #%d: %d\n",cnt,dis[n]);        else            printf("Case #%d: -1\n",cnt);    }    return 0;}