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

The Shortest Path in Nya Graph

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 37    Accepted Submission(s): 6

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

 

Recommend

zhuyuanchen520

 

最短路。

主要是建图。

N个点，然后有N层，要假如2*N个点。

总共是3*N个点。

点1~N就是对应的实际的点1~N.  要求的就是1到N的最短路。

然后点N+1 ~ 3*N 是N层拆出出来的点。

第i层，入边到N+2*i-1, 出边从N+2*i 出来。（1<= i <= N）

N + 2*i    到  N + 2*(i+1)-1 加边长度为C. 表示从第i层到第j层。

N + 2*(i+1) 到 N + 2*i - 1 加边长度为C,表示第i+1层到第j层。

如果点i属于第u层，那么加边 i -> N + 2*u -1         N + 2*u ->i  长度都为0

然后用优先队列优化的Dijkstra就可以搞出最短路了

 /* ***********************************************
 Author        :kuangbin
 Created Time  :2013-9-11 12:30:12
 File Name     :2013-9-11\1010.cpp
 ************************************************ */

 #include <stdio.h>
 #include <string.h>
 #include <iostream>
 #include <algorithm>
 #include <vector>
 #include <queue>
 #include <set>
 #include <map>
 #include <string>
 #include <math.h>
 #include <stdlib.h>
 #include <time.h>
 using namespace std;

 /*
  * 使用优先队列优化Dijkstra算法
  * 复杂度O(ElogE)
  * 注意对vector<Edge>E[MAXN]进行初始化后加边
  */
 const int INF=0x3f3f3f3f;
 const int MAXN=;
 struct qnode
 {
     int v;
     int c;
     qnode(int _v=,int _c=):v(_v),c(_c){}
     bool operator <(const qnode &r)const
     {
         return c>r.c;
     }
 };
 struct Edge
 {
     int v,cost;
     Edge(int _v=,int _cost=):v(_v),cost(_cost){}
 };
 vector<Edge>E[MAXN];
 bool vis[MAXN];
 int dist[MAXN];
 void Dijkstra(int n,int start)//点的编号从1开始
 {
     memset(vis,false,sizeof(vis));
     for(int i=;i<=n;i++)dist[i]=INF;
     priority_queue<qnode>que;
     while(!que.empty())que.pop();
     dist[start]=;
     que.push(qnode(start,));
     qnode tmp;
     while(!que.empty())
     {
         tmp=que.top();
         que.pop();
         int u=tmp.v;
         if(vis[u])continue;
         vis[u]=true;
         for(int i=;i<E[u].size();i++)
         {
             int v=E[tmp.v][i].v;
             int cost=E[u][i].cost;
             if(!vis[v]&&dist[v]>dist[u]+cost)
             {
                 dist[v]=dist[u]+cost;
                 que.push(qnode(v,dist[v]));
             }
         }
     }
 }
 void addedge(int u,int v,int w)
 {
     E[u].push_back(Edge(v,w));
 }

 int main()
 {
     //freopen("in.txt","r",stdin);
     //freopen("out.txt","w",stdout);
     int T;
     int N,M,C;
     scanf("%d",&T);
     int iCase = ;
     while(T--)
     {
         scanf("%d%d%d",&N,&M,&C);
         for(int i = ;i <= *N;i++) E[i].clear();
         int u,v,w;
         for(int i = ;i <= N;i++)
         {
             scanf("%d",&u);
             addedge(i,N + *u - ,);
             addedge(N + *u ,i,);

         }
         for(int i = ;i < N;i++)
         {
             addedge(N + *i-,N + *(i+),C);
             addedge(N + *(i+)-,N + *i,C);
         }
         while(M--)
         {
             scanf("%d%d%d",&u,&v,&w);
             addedge(u,v,w);
             addedge(v,u,w);
         }
         Dijkstra(*N,);
         iCase++;
         if(dist[N] == INF)dist[N] = -;
         printf("Case #%d: %d\n",iCase,dist[N]);

     }
     return ;
 }