Choose the best route

针对起点众多的最短路径问题,通过引入虚拟节点并运用Dijkstra算法实现高效求解。适用于城市交通路线规划等场景。

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Problem Description
One day , Kiki wants to visit one of her friends. As she is liable to carsickness , she wants to arrive at her friend’s home as soon as possible . Now give you a map of the city’s traffic route, and the stations which are near Kiki’s home so that she can take. You may suppose Kiki can change the bus at any station. Please find out the least time Kiki needs to spend. To make it easy, if the city have n bus stations ,the stations will been expressed as an integer 1,2,3…n.
 

 

Input
There are several test cases.
Each case begins with three integers n, m and s,(n<1000,m<20000,1=<s<=n) n stands for the number of bus stations in this city and m stands for the number of directed ways between bus stations .(Maybe there are several ways between two bus stations .) s stands for the bus station that near Kiki’s friend’s home.
Then follow m lines ,each line contains three integers p , q , t (0<t<=1000). means from station p to station q there is a way and it will costs t minutes .
Then a line with an integer w(0<w<n), means the number of stations Kiki can take at the beginning. Then follows w integers stands for these stations.
 

 

Output
The output contains one line for each data set : the least time Kiki needs to spend ,if it’s impossible to find such a route ,just output “-1”.
 

 

Sample Input
5 8 5 1 2 2 1 5 3 1 3 4 2 4 7 2 5 6 2 3 5 3 5 1 4 5 1 2 2 3 4 3 4 1 2 3 1 3 4 2 3 2 1 1
题意:给定n个点,m条路,一个目标位置,其实位置有多个,求一路最短路
题解:因为起点过多,如果对每一个起点进行djkstra算法会超时。采用增加一个虚拟结点的办法一遍djkstra即可
#include<stdio.h>
#include<iostream>
#include<string.h>
using namespace std;
const int inf=10000000;
int n,m,s;
int p[1001][1001];
int visit[1001];
int des[1001];
int dijkstra()
{
 int i,j,temp,k;
 memset(visit,0,sizeof(visit));
 for(i=0;i<=n;i++) des[i]=p[0][i];
 des[0]=0;
 visit[0]=1;
 for(i=0;i<=n;i++)
 {
  temp=inf;
  for(j=0;j<=n;j++)
  {
   if(visit[j]==0&&des[j]<temp)
   {
    temp=des[j];
    k=j;
   }
  }
  visit[k]=1;
  if(temp==inf) break;
  for(j=0;j<=n;j++)
  {
   if(visit[j]==0&&des[k]+p[k][j]<des[j]) des[j]=des[k]+p[k][j];
  }
 }
 return des[s];
}
int main()
{
 int a,b,c,mm,x,r,jk,i,j;
    while(scanf("%d %d %d",&n,&m,&s)!=EOF)
 {
  jk=10000000;
  for(i=0;i<=n;i++)
  {
   for(j=0;j<=n;j++)
   {
    p[i][j]=inf;
   }
  }
        for(i=1;i<=m;i++)
  {
   scanf("%d %d %d",&a,&b,&c);
   if(p[a][b]>c)
   {
    p[a][b]=c;
   }
  }
  scanf("%d",&mm);
  for(i=0;i<mm;i++)
  {
   scanf("%d",&x);
            p[0][x]=0;
  }
  jk=dijkstra();
  if(jk==10000000) puts("-1");
  else printf("%d\n",jk);
 }
 return 0;
}

转载于:https://www.cnblogs.com/ffhuguang/p/3288238.html

### 计算机网络基础:路由交换原理及区别 #### 一、路由的概念及其功能 路由器作为一种关键的网络设备,在计算机网络中承担着至关重要的角色。其核心职责在于跨不同网络间的数据包传输,确保数据能够准确无误地从源头抵达目的地,并且在此过程中寻找最优路径来提高效率和可靠性[^2]。 #### 二、交换的概念及其特性 相比之下,交换发生在同一局域网内部的不同节点之间。交换机依据MAC地址表快速查找目标端口并将帧发送出去;这种操作几乎是在硬件层面完成的,因此速度非常快。而且由于采用了全双工通信模式,使得在同一时刻可以实现双向的信息交流而不会发生冲突。 #### 三、两者的主要差异点 - **工作层次**:路由位于OSI七层模型中的第三层——即网络层,处理的是IP地址级别的逻辑寻址问题;而交换则处于第二层数据链路层,关注物理介质上的位流控制以及基于MAC地址的身份识别。 - **覆盖范围**:前者适用于广域网(WAN),后者多见于局域网(LAN)环境内; - **决策机制**:对于路由而言,选择最佳路径涉及到复杂的算法计算(如RIP协议按照定时更新的方式共享邻居间的距离矢量信息),并且可能跨越多个子网边界;相反,简单的查表匹配足以满足大多数情况下交换的需求[^3]。 ```python # Python伪代码展示如何模拟基本的路由选择过程 def choose_best_path(routes): best_route = None lowest_metric = float('inf') for route in routes: if route['metric'] < lowest_metric: lowest_metric = route['metric'] best_route = route return best_route routes_example = [ {'destination': '192.168.1.0/24', 'next_hop': '10.0.0.1', 'metric': 1}, {'destination': '172.16.0.0/16', 'next_hop': '10.0.0.2', 'metric': 5} ] print(choose_best_path(routes_example)) ```
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