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+10;
bool vis[MAXN];
int N, M;

struct Edge
{
    int nxt, no, v;
}edge[MAXN<<2];
int head[MAXN], tot;
void init()
{
    memset(vis, 0, sizeof(vis));
    memset(head, -1, sizeof(head));
    tot = 0;
}

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[1]; i != -1; i = edge[i].nxt){
        x.u = 1;
        x.v = edge[i].v;
        x.cnt = 1;
        x.type = edge[i].no;
        quq.push(x);
    }
    vis[1] = 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 != -1; i = edge[i].nxt){
            vv = edge[i].v;
            if(vis[vv]) continue;
            if(edge[i].no != t) y.cnt = sum+1;
            else y.cnt = sum;
            y.v = vv;
            y.type = edge[i].no;
            quq.push(y);
        }
    }
    return -1;
}

int main()
{
    int u, v, no;
    while(~scanf("%d %d", &N, &M)){
        init();
        for(int i = 1; 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 0;
}