HDU-6386-最短路

Age of Moyu

Time Limit: 5000/2500 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 119    Accepted Submission(s): 23

Problem Description

Mr.Quin love fishes so much and Mr.Quin’s city has a nautical system,consisiting of N ports and M shipping lines. The ports are numbered 1 to N. Each line is occupied by a Weitian. Each Weitian has an identification number.

The i-th (1≤i≤M) line connects port Ai and Bi (Ai≠Bi) bidirectionally, and occupied by Ci Weitian (At most one line between two ports).

When Mr.Quin only uses lines that are occupied by the same Weitian, the cost is 1 XiangXiangJi. Whenever Mr.Quin changes to a line that is occupied by a different Weitian from the current line, Mr.Quin is charged an additional cost of 1 XiangXiangJi. In a case where Mr.Quin changed from some Weitian A's line to another Weitian's line changes to Weitian A's line again, the additional cost is incurred again.

Mr.Quin is now at port 1 and wants to travel to port N where live many fishes. Find the minimum required XiangXiangJi (If Mr.Quin can’t travel to port N, print −1instead)

 

Input

There might be multiple test cases, no more than 20. You need to read till the end of input.

For each test case,In the first line, two integers N (2≤N≤100000) and M (0≤M≤200000), representing the number of ports and shipping lines in the city.

In the following m lines, each contain three integers, the first and second representing two ends Ai and Bi of a shipping line (1≤Ai,Bi≤N) and the third representing the identification number Ci (1≤Ci≤1000000) of Weitian who occupies this shipping line.

 

Output

For each test case output the minimum required cost. If Mr.Quin can’t travel to port N, output −1 instead.

 

Sample Input

3 3
1 2 1
1 3 2
2 3 1
2 0
3 2
1 2 1
2 3 2

 

Sample Output

1 -1 2　　

　　　　

　　　　题面贼鸡儿难理解，其实就是求最小换乘次数，第一次上路时也算一次换乘。用set维护当前最短路的状态对应的边权，然后跑最短路就好了。用spfa，不断更新，看讨论说bfs也可以做。dij的话也ok，只要不断更新所有的最短路状态就好了。

　　

 #include<bits/stdc++.h>
 using namespace std;
 #define LL long long
 #define mp make_pair
 #define pb push_back
 #define inf 0x3f3f3f3f
 int N,M;
 struct Edge{
     int v,w,next;
 }e[];
 int tot,first[],d[];
 bool vis[];
 set<int>S[];
 void add(int u,int v,int w){
     e[tot].v=v;
     e[tot].w=w;
     e[tot].next=first[u];
     first[u]=tot++;
 }

  　　 int spfa(){
     memset(vis,,sizeof(vis));
     memset(d,inf,sizeof(d));
     for(int i=;i<=N;++i)S[i].clear();
     queue<int>q;
     q.push();
     d[]=;
     vis[]=;
     while(!q.empty()){
         int u=q.front();
         q.pop();
         vis[u]=;
         for(int i=first[u];~i;i=e[i].next){
             int w;
             if(S[u].count(e[i].w)) w=d[u];
             else w=d[u]+;
             if(d[e[i].v]>w){
                 d[e[i].v]=w;
                 S[e[i].v].clear();
                 S[e[i].v].insert(e[i].w);
                 if(!vis[e[i].v]){
                     q.push(e[i].v);
                     vis[e[i].v]=;
                 }
             }
             else if(d[e[i].v]==w&&S[e[i].v].count(e[i].w)==){
                 S[e[i].v].insert(e[i].w);
                     if(!vis[e[i].v]){
                     q.push(e[i].v);
                     vis[e[i].v]=;
                 }
             }
         }
     }
     if(d[N]==inf) return -;
     return d[N];
 }
 int main(){
     int i,j,u,v,w;
     while(scanf("%d%d",&N,&M)!=EOF){
         memset(first,-,sizeof(first));
         tot=;
         for(i=;i<=M;++i){
             scanf("%d%d%d",&u,&v,&w);
             add(u,v,w);
             add(v,u,w);
         }
         cout<<spfa()<<endl;
     }
     return ;
 }

 

Source

2018 Multi-University Training Contest 7