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)
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.
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。
1 #include<bits/stdc++.h> 2 using namespace std; 3 #define LL long long 4 #define mp make_pair 5 #define pb push_back 6 #define inf 0x3f3f3f3f 7 int N,M; 8 struct Edge{ 9 int v,w,next; 10 }e[400040]; 11 int tot,first[101010],d[101010]; 12 bool vis[101010]; 13 set<int>S[101010]; 14 void add(int u,int v,int w){ 15 e[tot].v=v; 16 e[tot].w=w; 17 e[tot].next=first[u]; 18 first[u]=tot++; 19 }
int spfa(){ 27 memset(vis,0,sizeof(vis)); 28 memset(d,inf,sizeof(d)); 29 for(int i=1;i<=N;++i)S[i].clear(); 30 queue<int>q; 31 q.push(1); 32 d[1]=0; 33 vis[1]=1; 34 while(!q.empty()){ 35 int u=q.front(); 36 q.pop(); 37 vis[u]=0; 38 for(int i=first[u];~i;i=e[i].next){ 39 int w; 40 if(S[u].count(e[i].w)) w=d[u]; 41 else w=d[u]+1; 42 if(d[e[i].v]>w){ 43 d[e[i].v]=w; 44 S[e[i].v].clear(); 45 S[e[i].v].insert(e[i].w); 46 if(!vis[e[i].v]){ 47 q.push(e[i].v); 48 vis[e[i].v]=1; 49 } 50 } 51 else if(d[e[i].v]==w&&S[e[i].v].count(e[i].w)==0){ 52 S[e[i].v].insert(e[i].w); 53 if(!vis[e[i].v]){ 54 q.push(e[i].v); 55 vis[e[i].v]=1; 56 } 57 } 58 } 59 } 60 if(d[N]==inf) return -1; 61 return d[N]; 62 } 63 int main(){ 64 int i,j,u,v,w; 65 while(scanf("%d%d",&N,&M)!=EOF){ 66 memset(first,-1,sizeof(first)); 67 tot=0; 68 for(i=1;i<=M;++i){ 69 scanf("%d%d%d",&u,&v,&w); 70 add(u,v,w); 71 add(v,u,w); 72 } 73 cout<<spfa()<<endl; 74 } 75 return 0; 76 }
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