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[400040];
int tot,first[101010],d[101010];
bool vis[101010];
set<int>S[101010];
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,0,sizeof(vis));
    memset(d,inf,sizeof(d));
    for(int i=1;i<=N;++i)S[i].clear();
    queue<int>q;
    q.push(1);
    d[1]=0;
    vis[1]=1;
    while(!q.empty()){
        int u=q.front();
        q.pop();
        vis[u]=0;
        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]+1;
            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]=1;
                }
            }
            else if(d[e[i].v]==w&&S[e[i].v].count(e[i].w)==0){
                S[e[i].v].insert(e[i].w);
                    if(!vis[e[i].v]){
                    q.push(e[i].v);
                    vis[e[i].v]=1;
                }
            }
        }
    }
    if(d[N]==inf) return -1;    
    return d[N];
}
int main(){
    int i,j,u,v,w;
    while(scanf("%d%d",&N,&M)!=EOF){
        memset(first,-1,sizeof(first));
        tot=0;
        for(i=1;i<=M;++i){
            scanf("%d%d%d",&u,&v,&w);
            add(u,v,w);
            add(v,u,w);
        }
        cout<<spfa()<<endl;
    }
    return 0;
}

 

Source

2018 Multi-University Training Contest 7