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 <= 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

 

 

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

 

 

  1 /* ***********************************************

  2 Author        :kuangbin

  3 Created Time  :2013-9-11 12:30:12

  4 File Name     :2013-9-11\1010.cpp

  5 ************************************************ */

  6 

  7 #include <stdio.h>

  8 #include <string.h>

  9 #include <iostream>

 10 #include <algorithm>

 11 #include <vector>

 12 #include <queue>

 13 #include <set>

 14 #include <map>

 15 #include <string>

 16 #include <math.h>

 17 #include <stdlib.h>

 18 #include <time.h>

 19 using namespace std;

 20 

 21 /*

 22  * 使用优先队列优化Dijkstra算法

 23  * 复杂度O(ElogE)

 24  * 注意对vector<Edge>E[MAXN]进行初始化后加边

 25  */

 26 const int INF=0x3f3f3f3f;

 27 const int MAXN=1000010;

 28 struct qnode

 29 {

 30     int v;

 31     int c;

 32     qnode(int _v=0,int _c=0):v(_v),c(_c){}

 33     bool operator <(const qnode &r)const

 34     {

 35         return c>r.c;

 36     }

 37 };

 38 struct Edge

 39 {

 40     int v,cost;

 41     Edge(int _v=0,int _cost=0):v(_v),cost(_cost){}

 42 };

 43 vector<Edge>E[MAXN];

 44 bool vis[MAXN];

 45 int dist[MAXN];

 46 void Dijkstra(int n,int start)//点的编号从1开始

 47 {

 48     memset(vis,false,sizeof(vis));

 49     for(int i=1;i<=n;i++)dist[i]=INF;

 50     priority_queue<qnode>que;

 51     while(!que.empty())que.pop();

 52     dist[start]=0;

 53     que.push(qnode(start,0));

 54     qnode tmp;

 55     while(!que.empty())

 56     {

 57         tmp=que.top();

 58         que.pop();

 59         int u=tmp.v;

 60         if(vis[u])continue;

 61         vis[u]=true;

 62         for(int i=0;i<E[u].size();i++)

 63         {

 64             int v=E[tmp.v][i].v;

 65             int cost=E[u][i].cost;

 66             if(!vis[v]&&dist[v]>dist[u]+cost)

 67             {

 68                 dist[v]=dist[u]+cost;

 69                 que.push(qnode(v,dist[v]));

 70             }

 71         }

 72     }

 73 }

 74 void addedge(int u,int v,int w)

 75 {

 76     E[u].push_back(Edge(v,w));

 77 }

 78 

 79 int main()

 80 {

 81     //freopen("in.txt","r",stdin);

 82     //freopen("out.txt","w",stdout);

 83     int T;

 84     int N,M,C;

 85     scanf("%d",&T);

 86     int iCase = 0;

 87     while(T--)

 88     {

 89         scanf("%d%d%d",&N,&M,&C);

 90         for(int i = 1;i <= 3*N;i++) E[i].clear();

 91         int u,v,w;

 92         for(int i = 1;i <= N;i++)

 93         {

 94             scanf("%d",&u);

 95             addedge(i,N + 2*u - 1,0);

 96             addedge(N + 2*u ,i,0);

 97 

 98         }

 99         for(int i = 1;i < N;i++)

100         {

101             addedge(N + 2*i-1,N + 2*(i+1),C);

102             addedge(N + 2*(i+1)-1,N + 2*i,C);

103         }

104         while(M--)

105         {

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

107             addedge(u,v,w);

108             addedge(v,u,w);

109         }

110         Dijkstra(3*N,1);

111         iCase++;

112         if(dist[N] == INF)dist[N] = -1;

113         printf("Case #%d: %d\n",iCase,dist[N]);

114 

115     }

116     return 0;

117 }