HDU 4725 The Shortest Path in Nya Graph

he Shortest Path in Nya Graph

Time Limit: 1000ms

Memory Limit: 32768KB

This problem will be judged on HDU. Original ID: 4725
64-bit integer IO format: %I64d Java class name: Main

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

Sample Output

Case #1: 2

Case #2: 3

Source

2013 ACM/ICPC Asia Regional Online —— Warmup2

解题：

瞎搞搞。。

 #include <bits/stdc++.h>

 #define pii pair<int,int>

 using namespace std;

 const int INF = 0x3f3f3f3f;

 const int maxn = ;

 struct arc{

     int to,w,next;

     arc(int x = ,int y = ,int z = -){

         to = x;

         w = y;

         next = z;

     }

 }e[];

 int head[maxn],d[maxn],tot,n,m,c;

 int layer[maxn];

 void add(int u,int v,int w){

     e[tot] = arc(v,w,head[u]);

     head[u] = tot++;

 }

 bool done[maxn];

 priority_queue< pii,vector< pii >,greater< pii > >q;

 int dijkstra(int s,int t){

     while(!q.empty()) q.pop();

     memset(d,0x3f,sizeof d);

     memset(done,false,sizeof done);

     q.push(pii(d[s] = ,s));

     while(!q.empty()){

         int u = q.top().second;

         q.pop();

         if(done[u]) continue;

         done[u] = true;

         for(int i = head[u]; ~i; i = e[i].next){

             if(d[e[i].to] > d[u] + e[i].w){

                 d[e[i].to] = d[u] + e[i].w;

                 q.push(pii(d[e[i].to],e[i].to));

             }

         }

     }

     return d[t] == INF?-:d[t];

 }

 bool hslv[maxn];

 int main(){

     int kase,tmp,u,v,w,cs = ;

     scanf("%d",&kase);

     while(kase--){

         memset(head,-,sizeof head);

         memset(hslv,false,sizeof hslv);

         tot = ;

         scanf("%d%d%d",&n,&m,&c);

         for(int i = ; i <= n; ++i){

             scanf("%d",&tmp);

             layer[i] = tmp;

             hslv[tmp] = true;

         }

         for(int i = ; i < m; ++i){

             scanf("%d%d%d",&u,&v,&w);

             add(u,v,w);

             add(v,u,w);

         }

         for(int i = ; i <= n; ++i){

             add(layer[i]+n,i,);

             if(layer[i] > ) add(i,layer[i]-+n,c);

             if(layer[i] < n) add(i,layer[i]+n+,c);

         }

         printf("Case #%d: %d\n",cs++,dijkstra(,n));

     }

     return ;

 }