HDU 6386 Age of Moyu

 

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

 

题解：

最短路的写法；始终去找最小费用，保存上一次的边权和这一次比较，然后查看应当+1还是+0

我是根据https://blog.csdn.net/m0_37611893/article/details/81636688 这个链接而写，

#include <bits/stdc++.h>
using namespace std;
const int MAXN=100010;
const int INF=0x3f3f3f3f;
int n,m;
struct qnode{
    int v;   //下一个点
    int c;   //当前花费
    int w;   //当前这条边的价值
    qnode(int _v = 0, int _c = 0, int _w = 0):v(_v),c(_c),w(_w){}
    bool operator < (const qnode &r) const{    //按照费用小的在前
        return c>r.c;
    }
};
struct edge{
    int v,cost;
    edge(int _v=0,int _cost=0):v(_v),cost(_cost){}
};
vector<edge>E[MAXN];
int dis[MAXN];
int vis[MAXN];
void dij()
{
    for (int i = 0; i <=n ; ++i) {
        dis[i]=INF;
    }
    memset(vis, false, sizeof(vis));
    priority_queue<qnode>q;
    while(!q.empty()) q.pop();
    dis[1]=0;
    q.push(qnode(1,0,0));
    qnode tmp;
    while (!q.empty())
    {
        tmp=q.top();q.pop();
        int u=tmp.v;
        if(vis[u]) continue;
        vis[u]=true;
        dis[u]=tmp.c;
        for (int i = 0; i <E[u].size() ; ++i) {
            int v=E[u][i].v;
            int w1=E[u][i].cost;
            if(!vis[v])
            {
                int temp;
                if(tmp.w!=w1) temp=tmp.c+1;//如果这个边的值和上一条边值不一样，费用+1;
                else temp=tmp.c;
                q.push(qnode(v,temp,w1));
            }
        }
    }

}
int main()
{
    while (scanf("%d%d",&n,&m)!=EOF)
    {
        for (int i = 0; i <=n ; ++i) {
            E[i].clear();
        }
        int a,b,c;
        for (int i = 0; i <m ; ++i) {
            scanf("%d%d%d",&a,&b,&c);
            E[a].push_back(edge(b,c));
            E[b].push_back(edge(a,c));
        }
        dij();
        if(dis[n]==INF) printf("-1\n");
        else printf("%d\n",dis[n]);
    }
    return 0;
}