HDOJ 题目4725 The Shortest Path in Nya Graph（spfa，建图）

The Shortest Path in Nya Graph

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2824    Accepted Submission(s): 668

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

23 3 31 3 21 2 12 3 11 3 33 3 31 3 21 2 22 3 21 3 4

 

Sample Output

Case #1: 2Case #2: 3

 

Source

2013 ACM/ICPC Asia Regional Online —— Warmup2

 

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 ac代码

#include<stdio.h>#include<string.h>#include<stdlib.h>#include<queue>#include<iostream>#define INF 0xfffffffusing namespace std;struct s{int u,v,w,next;}edge[2000010];int n,m,c,vis[200100],lay[200100],head[200100],dis[200100],cnt;void add(int u,int v,int w){edge[cnt].u=u;edge[cnt].v=v;edge[cnt].w=w;edge[cnt].next=head[u];head[u]=cnt++;}void spfa(){int i;memset(vis,0,sizeof(vis));for(i=1;i<=n+n;i++)dis[i]=INF;queue<int>q;dis[1]=0;vis[1]=1;q.push(1);while(!q.empty()){int u=q.front();q.pop();vis[u]=0;for(i=head[u];i!=-1;i=edge[i].next){int v=edge[i].v;if(dis[v]>dis[u]+edge[i].w){dis[v]=dis[u]+edge[i].w;if(!vis[v]){q.push(v);vis[v]=1;}}}}}int main(){int t,cot=0;scanf("%d",&t);while(t--){int i;cnt=0;scanf("%d%d%d",&n,&m,&c);memset(vis,0,sizeof(vis));for(i=1;i<=n;i++){int x;scanf("%d",&x);lay[i]=x;vis[x]=1;}memset(head,-1,sizeof(head));for(i=1;i<n;i++){if(vis[i]&&vis[i+1]){add(i+n,n+i+1,c);add(i+n+1,i+n,c);}}for(i=1;i<=n;i++){add(n+lay[i],i,0);if(lay[i]>1)add(i,n+lay[i]-1,c);if(lay[i]<n)add(i,n+lay[i]+1,c);}for(i=1;i<=m;i++){int u,v,w;scanf("%d%d%d",&u,&v,&w);add(u,v,w);add(v,u,w);}spfa();printf("Case #%d: ",++cot);if(dis[n]==INF)printf("-1\n");elseprintf("%d\n",dis[n]);}}