Choose the best route

最新推荐文章于 2022-08-29 17:13:17 发布

转载最新推荐文章于 2022-08-29 17:13:17 发布 · 137 阅读

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原文链接：http://www.cnblogs.com/ffhuguang/p/3288238.html

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#数据结构与算法

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;
}