HDU 4725 The Shortest Path in Nya Graph （最短路 ）

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

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

分析：

本题的坑点在于建图的时候边数太多，因为每x层与x+1层中的每一个点都相互连接，这样就有 Nx^N(x+1) 条边，sum（Ni）为10^5，暴力建图肯定会爆的

建图的时候，每一层增加一个“层点”作为中介，然后层点与层点建边，层点与在该层上的点建边（边长为0），点与相邻层点建边（边长为c）。

最后增加的边，点与点建边。

在赛场上怂了，看见10^5没敢多建边，只想着如何在层之间建边了。。。对时间复杂度与时间限制的把握还是不好，以后要记住常用算法的复杂度，还要学习如何准确分析时间复杂度

代码：

  1 #include <iostream>
  2 #include <cstdio>
  3 #include <cstring>
  4 #include <queue>
  5 #include <vector>
  6 #define maxn 200005
  7 #define INF 0x3f3f3f3f
  8 using namespace std;
  9
 10 int n,m,c,ans,cnt;
 11 bool vis[maxn],vv[maxn];
 12 int dist[maxn],pp[maxn],lay[maxn];
 13 struct Node
 14 {
 15     int v,w;
 16     int next;
 17 } edge[20*maxn];
 18 queue<int>q;
 19
 20 void init()
 21 {
 22     memset(pp,0,sizeof(pp));
 23     memset(vv,0,sizeof(vv));
 24     memset(dist,0x3f,sizeof(dist));
 25 }
 26 void addedge(int u,int v,int w)
 27 {
 28     cnt++;
 29     edge[cnt].v=v;
 30     edge[cnt].w=w;
 31     edge[cnt].next=pp[u];
 32     pp[u]=cnt;
 33 }
 34 void SPFA()
 35 {
 36     int i,j,u,v,w;
 37     memset(vis,0,sizeof(vis));
 38     while(!q.empty()) q.pop();
 39     dist[1]=0;
 40     vis[1]=1;
 41     q.push(1);
 42     while(!q.empty())
 43     {
 44         u=q.front();
 45         q.pop();
 46         vis[u]=0;
 47         for(i=pp[u]; i; i=edge[i].next)
 48         {
 49             v=edge[i].v;
 50             w=edge[i].w;
 51             if(dist[v]>dist[u]+w)
 52             {
 53                 dist[v]=dist[u]+w;
 54                 if(!vis[v])
 55                 {
 56                     vis[v]=1;
 57                     q.push(v);
 58                 }
 59             }
 60         }
 61     }
 62 }
 63 int main()
 64 {
 65     int i,j,t,u,v,w,test=0;
 66     scanf("%d",&t);
 67     while(t--)
 68     {
 69         scanf("%d%d%d",&n,&m,&c);
 70         init();
 71         cnt=0;
 72         for(i=1; i<=n; i++)
 73         {
 74             scanf("%d",&u);
 75             lay[i]=u;
 76             vv[u]=1;
 77         }
 78         for(i=1; i<n; i++)
 79         {
 80             if(vv[i]&&vv[i+1])  // 两层都出现过点相邻层才建边
 81             {
 82                 addedge(n+i,n+i+1,c);
 83                 addedge(n+i+1,n+i,c);
 84             }
 85         }
 86         for(i=1; i<=n; i++)     // 层到点建边  点到相邻层建边
 87         {
 88             addedge(n+lay[i],i,0);
 89             if(lay[i]>1) addedge(i,n+lay[i]-1,c);
 90             if(lay[i]<n) addedge(i,n+lay[i]+1,c);
 91         }
 92         for(i=1; i<=m; i++)     // 点到点建边
 93         {
 94             scanf("%d%d%d",&u,&v,&w);
 95             addedge(u,v,w);
 96             addedge(v,u,w);
 97         }
 98         SPFA();
 99         printf("Case #%d: ",++test);
100         ans=dist[n];
101         if(ans<INF) printf("%d\n",ans);
102         else printf("-1\n");
103     }
104     return 0;
105 }