Age of Moyu

Time Limit: 5000/2500 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 4523 Accepted Submission(s): 430

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

题意：n个港口和m条边，没条边都会占用一条航线，航线标号为c，每次航线的变换都会花费1元，求1到n的最小花费。

广搜了一遍，但是花费相同到next点的航线不同也需要进队，这样即可包含所有情况了。

#include<stdio.h>
#include<algorithm>
#include<vector>
#include<queue>
using namespace std;
#define ll long long
const int maxm = 200005;
const int INF = 1e9 + 7;
struct node
{
    int v, c;
};
vector<node>v[maxm];
int n, dis[maxm], vis[maxm], p[maxm], lc[maxm];
void bfs()
{
    queue<int>q;
    for (int i = 1;i <= n;i++) dis[i] = INF, lc[i] = -1;
    dis[1] = 0, vis[1] = 1;
    q.push(1);
    while (!q.empty())
    {
        int u = q.front();q.pop();
        for (int i = 0;i < v[u].size();i++)
        {
            node now = v[u][i];
            int len = 0;
            if (lc[u] != now.c) len = 1;
            //if (u == 2 && now.v == 3) printf("%d %d %d\n", dis[u], dis[now.v],len);
            if (dis[now.v] > dis[u] + len || dis[now.v] == dis[u] + len&&lc[now.v] != now.c)
            {
                lc[now.v] = now.c;
                dis[now.v] = dis[u] + len;
                q.push(now.v);
            }
        }
    }
}
int main()
{
    int i, j, k, sum, t, m, c, x, y;
    while (scanf("%d%d", &n, &m)!=EOF)
    {
        for (i = 1;i <= n;i++) v[i].clear();
        for (i = 1;i <= m;i++)
        {
            scanf("%d%d%d", &x, &y, &c);
            node tmp;tmp.c = c;
            tmp.v = y, v[x].push_back(tmp);
            tmp.v = x, v[y].push_back(tmp);
        }
        bfs();
        if (dis[n] == INF) printf("-1\n");
        else printf("%d\n", dis[n]);
    }
    return 0;
}

2018 Multi-University Training Contest 7 1001 Age of Moyu

Age of Moyu

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