hdu6386 Age of Moyu【最短路】

Age of Moyu

Time Limit: 5000/2500 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 3125    Accepted Submission(s): 981

 

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

 

 

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其实当时没怎么认真思考这道题要怎么写

补题时候看的题解

有个题解说用set 其实没怎么搞清楚要怎么写 但是这种方法也可以实现

用优先队列优化dijkstra 队列维护的是边而不是点

根据当前边的编号决定cost是否增加

因为是优先队列所以可以保证最先到达n的线路是cost最小的

好像比赛的时候他们写T了好多次....不知道他们是怎么写的

 

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<stack>
#include<queue>
#define inf 0x3f3f3f3f
using namespace std;

int n, m;
const int maxn = 100005;
const int maxm = 200005;
int head[maxn], cnt;
struct edge{
    int v, f, w;
    int nxt;
    bool operator < (edge a) const{
        return w > a.w;
    }
}e[maxm * 2];

void addedge(int u, int v, int f)
{
    e[cnt].v = v;
    e[cnt].f = f;
    e[cnt].nxt = head[u];
    head[u] = cnt;
    cnt++;
}

int dis[maxn], vis[maxn];
void init()
{
    cnt = 0;
    memset(vis, 0, sizeof(vis));
    memset(dis, inf, sizeof(dis));
    memset(head, -1, sizeof(head));
}

void dijkstra()
{
    dis[1] = 0;
    priority_queue <edge> q;
    while(!q.empty()){
        q.pop();
    }
    edge now;
    now.v = 1;
    now.f = 0;
    now.w = 0;
    q.push(now);
    while(!q.empty()){
        now = q.top();
        q.pop();
        int u = now.v;
        if(vis[u]){
            continue;
        }
        vis[u] = true;
        dis[u] = now.w;
        for(int i = head[u]; ~i; i = e[i].nxt){
            int v = e[i].v;
            if(!vis[v]){
                edge nex;
                nex.v = v;
                nex.f = e[i].f;
                nex.w = dis[u] + (e[i].f != now.f);
                q.push(nex);
            }
        }
    }
}

int main()
{
	while(scanf("%d%d", &n, &m) != EOF){
        init();

        for(int i = 0; i < m; i++){
            int a, b, c;
            scanf("%d%d%d", &a, &b, &c);
            addedge(a, b, c);
            addedge(b, a, c);
        }

        dijkstra();
        if(dis[n] == inf){
            printf("-1\n");
        }
        else{
            printf("%d\n", dis[n]);
        }
	}
	return 0;
}