hdu 4725

The Shortest Path in Nya Graph

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2296    Accepted Submission(s): 561

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

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

 

Source

2013 ACM/ICPC Asia Regional Online —— Warmup2

 

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最短路

把每一层拆成2 个点 一个是连接结点入边，一个是连接结点出边，每层连接结点入边的点连接下个层连接结点出边的点，下一层连接入边的点连接上一层连接出边的点。

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 #include <vector>
 #include <queue>

 using namespace std;

 const int MAX_N = 3e5 + ;
 const int INF = 1e9 + ;
 typedef long long ll;
 struct heapnode {
         int d;
         int  u;
         bool operator < (const heapnode &rhs) const {
                 return d > rhs.d;
         }
 };
 struct Edge {int from, to, cost;};

 vector<int> G[MAX_N];
 vector<Edge> edges;
 int N,M,C;
 int d[MAX_N];
 bool done[MAX_N];
 vector <int> lay[MAX_N];

 void add_edge(int from, int to, int cost) {
         edges.push_back((Edge){from, to, cost});
         int m = edges.size();
         G[from].push_back(m - );
 }

 void dij(int s) {
         memset(done,,sizeof(done));
         for(int i = ; i <=  * N; ++i) {
                 d[i] = INF;
         }
         d[s] = ;
         priority_queue <heapnode> q;
         q.push((heapnode) {d[s], s});

         while(!q.empty()) {
                 heapnode x = q.top(); q.pop();
                 int u = x.u;
                 if(done[u]) continue;
                 done[u] = ;
                 if(u == N) return;

                 for(int i = ; i < G[u].size(); ++i) {
                         Edge &e = edges[ G[u][i] ];
                         if(d[e.to] > d[u] + e.cost) {
                                 d[e.to] = d[u] + e.cost;
                                 q.push((heapnode) {d[e.to], e.to});
                         }
                 }
         }
 }
 int main()
 {

    // freopen("sw.in","r",stdin);
     int t;
     scanf("%d",&t);
     int ca = ;
     while(t--) {
             scanf("%d%d%d",&N,&M,&C);
             for(int i = ; i <=  * N; ++i) G[i].clear();
             edges.clear();

             for(int i = ; i <= N; ++i) {
                     int ch;
                     scanf("%d",&ch);
                     add_edge(i, ch + N, );
                     add_edge(ch +  * N, i, );

             }

             /*for(int i = 2 * N + 1; i <= 3 * N; ++i) {
                     add_edge(i, i - N, 0);
             }*/

             for(int i = N + ; i <=  * N - ; ++i) {
                         add_edge(i, i +  + N, C);
                         add_edge(i + , i + N, C);
             }

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

             dij();
             printf("Case #%d: %d\n",ca++, (d[N] == INF) ? - : d[N]);
     }

     return ;
 }