HDU 6386 Age of Moyu 【BFS + 优先队列优化】

任意门：http://acm.hdu.edu.cn/showproblem.php?pid=6386

Age of Moyu

Time Limit: 5000/2500 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 3821    Accepted Submission(s): 1214

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

 

Source

2018 Multi-University Training Contest 7

题意概括：

N个点，M条无向边，每条边都有编号，经过相同的编号的边不会改变花费，经过不同的编号的边花费加一。

解题思路：

BFS 求最短路，用边替换点进队，按照花费升序排序。

可能数据偏水，可以卡过去。

AC code：

 #include <bits/stdc++.h>
 #define INF 0x3f3f3f3f
 #define LL long long
 using namespace std;

 const int MAXN = 1e5+;
 bool vis[MAXN];
 int N, M;

 struct Edge
 {
     int nxt, no, v;
 }edge[MAXN<<];
 int head[MAXN], tot;
 void init()
 {
     memset(vis, , sizeof(vis));
     memset(head, -, sizeof(head));
     tot = ;
 }

 void add(int a, int b, int no)
 {
     edge[tot].v = b;
     edge[tot].no = no;
     edge[tot].nxt = head[a];
     head[a] = tot++;
 }

 struct data
 {
     int u, v, cnt, type;
     bool operator < (const data &a) const {
         return cnt > a.cnt;
     }
 };

 int solve()
 {
     data x, y;
     int from, to, t, vv, sum;
     priority_queue<data>quq;
     for(int i = head[]; i != -; i = edge[i].nxt){
         x.u = ;
         x.v = edge[i].v;
         x.cnt = ;
         x.type = edge[i].no;
         quq.push(x);
     }
     vis[] = true;
     while(!quq.empty()){
         x = quq.top();
         quq.pop();
         to = x.v;
         t = x.type;
         sum = x.cnt;
         if(to == N) return sum;
         vis[to] = true;
         for(int i = head[to]; i != -; i = edge[i].nxt){
             vv = edge[i].v;
             if(vis[vv]) continue;
             if(edge[i].no != t) y.cnt = sum+;
             else y.cnt = sum;
             y.v = vv;
             y.type = edge[i].no;
             quq.push(y);
         }
     }
     return -;
 }

 int main()
 {
     int u, v, no;
     while(~scanf("%d %d", &N, &M)){
         init();
         for(int i = ; i <= M; i++){
             scanf("%d %d %d", &u, &v, &no);
             add(u, v, no);
             add(v, u, no);
         }

         int ans = solve();
         printf("%d\n", ans);
     }
     return ;
 }