HDU 6386 Age of Moyu（最短路变形）

Age of Moyu

Time Limit: 5000/2500 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 1531    Accepted Submission(s): 467

 

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

题意：给你一个图，每条边有一种颜色。你每走一种颜色消耗1分钟，然后沿该颜色的边继续走就不消耗时间，走其他颜色的边就会再消耗1分钟。求从1点走到n点至少需要几分钟，如果走不到n点输出-1。

思路：就是最短路的变形。虽然这道题好像标程错了还是原题（但是出一道题毕竟不容易）。。。有多种解法，这里我只说我比赛的时候过了的解法。

由于保证全局最优，我们每次只走一步，然后暴力dfs遍历这一步能走到的所有点（这样一定保证1点到每一个点都是最短的）。由于所有边的点都只走一次，保证了时间复杂度不会超过O(n+m)自己xjb乱算的，为了节省时间到达n直接返回。

代码：

#pragma comment(linker,"/STACK:10240000,10240000")
#include
#define ll long long
#define inf 0x3f3f3f3f
using namespace std;
const int maxn=400010;
struct node
{
    int id,u,v,cost;
    node(int _id=0,int _u=0,int _v=0,int _cost=0):id(_id),u(_u),v(_v),cost(_cost){}
};
queueq;
vectore[maxn];
int n,m,k;
int dis[maxn],vis[maxn];
void addedge(int id,int u,int v,int w)
{
    e[u].push_back(node(id,u,v,w));
}
void dfs(int x,int id,int dep,int fa)
{
    if(!vis[x]){q.push(x);dis[x]=dep;vis[x]=1;}
    if(dis[n]!=inf)return;
    for(int i=0;i