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

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+;
bool vis[MAXN];
int N, M; struct Edge
{
int nxt, no, v;
}edge[MAXN<<];
int head[MAXN], tot;
void init()
{
memset(vis, , sizeof(vis));
memset(head, -, sizeof(head));
tot = ;
} 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[]; i != -; i = edge[i].nxt){
x.u = ;
x.v = edge[i].v;
x.cnt = ;
x.type = edge[i].no;
quq.push(x);
}
vis[] = 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 != -; i = edge[i].nxt){
vv = edge[i].v;
if(vis[vv]) continue;
if(edge[i].no != t) y.cnt = sum+;
else y.cnt = sum;
y.v = vv;
y.type = edge[i].no;
quq.push(y);
}
}
return -;
} int main()
{
int u, v, no;
while(~scanf("%d %d", &N, &M)){
init();
for(int i = ; 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 ;
}

HDU 6386 Age of Moyu 【BFS + 优先队列优化】的更多相关文章

  1. HDU - 6386 Age of Moyu (双端队列+bfs)

    题目链接 双端队列跑边,颜色相同的边之间的花费为0,放进队首:不同的花费为1,放进队尾. 用Dijkstra+常数优化也能过 #include<bits/stdc++.h> using n ...

  2. HDU - 6386 Age of Moyu 2018 Multi-University Training Contest 7 (Dijkstra变型)

    题意:N个点M条边的无向图,每条边都有属于自己的编号,如果一条路径上的边编号都相同,那么花费仅为1:改变至不同编号的路径,花费加1,无论这个编号之前是否走过. 分析:记录每个点的最小花费,再用set维 ...

  3. hdu 6386 Age of Moyu (重边判断)

    本来用一个map判重边结果T了, 实际上可以直接给边上打标记即可 int n, m; struct _ {int to,w,vis;}; vector<_> g[N]; int dis[N ...

  4. HDU 6386 Age of Moyu

    Problem Description Mr.Quin love fishes so much and Mr.Quin’s city has a nautical system,consisiting ...

  5. HDU 6386 Age of Moyu (最短路+set)

    <题目链接> 题目大意:给定一张无向图,有n个点m条边,从一条边到另一条边,如果两边的指不同 花费就要+1,如果相同就不需要花费. 先从1走到n问最小花费是多少.(第一条边的花费都是1) ...

  6. hdu - 1180 诡异的楼梯 (bfs+优先队列)

    http://acm.hdu.edu.cn/showproblem.php?pid=1180 注意点就是楼梯是在harry移动完之后才会改变方向,那么只要统计到达这个点时间奇偶性,就可以知道当前楼梯是 ...

  7. HDU 6395 分段矩阵快速幂 HDU 6386 建虚点+dij

    http://acm.hdu.edu.cn/showproblem.php?pid=6395 Sequence Time Limit: 4000/2000 MS (Java/Others)    Me ...

  8. HDU 2822 (BFS+优先队列)

    题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=2822 题目大意:X消耗0,.消耗1, 求起点到终点最短消耗 解题思路: 每层BFS的结点,优先级不同 ...

  9. HDU 1242 -Rescue (双向BFS)&amp;&amp;( BFS+优先队列)

    题目链接:Rescue 进度落下的太多了,哎╮(╯▽╰)╭,渣渣我总是埋怨进度比别人慢...为什么不试着改变一下捏.... 開始以为是水题,想敲一下练手的,后来发现并非一个简单的搜索题,BFS做肯定出 ...

随机推荐

  1. MySQL触发器基本介绍

    基本简介: 1.触发器可以让你在执行insert,update,delete语句的时候,执行一些特定的操作.并且可以在MySQL中指定是在sql语句执行前触发还是执行后触发. 2.触发器没有返回值. ...

  2. 自动收缩数据库T-SQL

    alter database 数据库名 set auto_update_statistics off alter database 数据库名 set auto_update_statistics on

  3. Struts2 知识点梳理

    一.Struts2简介 1.概念:轻量级的MVC框架,主要解决了请求分发的问题,重心在控制层和表现层.低侵入性,与业务代码的耦合度很低.Struts2实现了MVC,并提供了一系列API,采用模式化方式 ...

  4. 三:Mybatis知识整理(3)

    一:mybatis中模糊查询的方法: 1.直接传参法:在java传参时进行拼接 -- %keyword% 2.mysql内置函数:concart('%',#{keyword},'%') -- 拼接sq ...

  5. 固态硬盘SSD与闪存(Flash Memory)

    转自:http://qiaodahai.com/solid-state-drives-ssd-and-flash-memory.html 固态硬盘SSD(Solid State Drive)泛指使用N ...

  6. How to limit Dialog's max height?

    1. We can make it to play trick in code. At Dialog's show function, after app has set contentView, w ...

  7. Thrift笔记(七)--回调源码分析

    网上找了写代码,东拼西凑写了个demo.开始server用的是阻塞io,不行,换成非阻塞的io就可以.这里可能需要注意下 thrift文件 namespace java com.gxf.thrift ...

  8. Java测试工具使用(1)--Junit

    在进行测试之前需要导入junit的两个包,分别是 junit:4.12;hamcrest-core:1.1 1.基本测试标签 @Test.@Before.@After 2.组测试 有时候多个测试文件, ...

  9. jQuery Text-to-Speech 谷歌在线语音

    <!DOCTYPE html> <html> <head> <meta content="text/html; charset=utf-8" ...

  10. 2018.10.26NOIP模拟赛解题报告

    心路历程 预计得分:\(100 + 100 + 70\) 实际得分:\(40 + 100 + 70\) 妈妈我又挂分了qwq..T1过了大样例就没管,直到临考试结束前\(10min\)才发现大样例是假 ...