hdu The Shortest Path in Nya Graph（建图+spfa）

 The Shortest Path in Nya Graph

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 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 <= 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

23 3 31 3 21 2 12 3 11 3 33 3 31 3 21 2 22 3 21 3 4

 

Sample Output

Case #1: 2Case #2: 3

ps：这道题主要就是难在建图上了，要将层抽象出来成为n个点（对应编号依次为n+1~n+n），然后层与层建边（相邻层都有点才能建边），点与点建边，层与该层上的点建边（边长为0），点与相邻层建边（边长为c）

代码：

#include<stdio.h>#include<string.h>#include<queue>using namespace std;#define maxn 200000+10#define maxv 800000+10#define mem(a,b)  memset(a,b,sizeof(a))const int inf=0x3f3f3f3f;struct node{    int v,w,next;} G[maxv];int d[maxn],first[maxn],lay[maxn];bool inq[maxn];int n,m,len;queue<int>q;void spfa(){    int i,st;    d[1]=0,inq[1]=1;    while(!q.empty())        q.pop();    q.push(1);    while(!q.empty())    {        st=q.front();        q.pop();        inq[st]=0;        for(i=first[st]; i!=-1; i=G[i].next)        {            int v=G[i].v,w=G[i].w;            if(d[v]>d[st]+w)            {                d[v]=d[st]+w;                if(!inq[v])                {                    inq[v]=1;                    q.push(v);                }            }        }    }}void add_egde(int u,int v,int w){    G[len].v=v,G[len].w=w;    G[len].next=first[u];    first[u]=len++;}int main(){    int t,k=1,c;    scanf("%d",&t);    while(k<=t)    {        mem(first,-1);        mem(lay,0);        mem(inq,0);        scanf("%d%d%d",&n,&m,&c);        len=1;        int i,u,v,w,level;        for(i=1;i<=n;i++)        {            scanf("%d",&level);            lay[i]=level;            inq[level]=1;        }        for(i=1; i<n; i++)        {            if(inq[i]&&inq[i+1])//两层都出现过点，相邻层才建边            add_egde(i+n,i+n+1,c);            add_egde(i+n+1,i+n,c);        }        for(i=1; i<=n; i++)//层到点建边   点到相邻层建边        {            d[i]=d[i+n]=inf;            inq[i]=inq[i+n]=0;            add_egde(n+lay[i],i,0);            if(lay[i]>1)                add_egde(i,lay[i]+n-1,c);            if(lay[i]<n)                add_egde(i,lay[i]+n+1,c);        }        for(i=1; i<=m; i++)//点到点建边        {            scanf("%d%d%d",&u,&v,&w);            add_egde(u,v,w);            add_egde(v,u,w);        }        spfa();        if(d[n]==inf||n==0)            printf("Case #%d: -1\n",k++);        else            printf("Case #%d: %d\n",k++,d[n]);    }    return 0;}