zoukankan      html  css  js  c++  java
  • HDU1595find the longest of the shortestdijkstra+记录路径

    Marica is very angry with Mirko because he found a new girlfriend and she seeks revenge.Since she doesn't live in the same city, she started preparing for the long journey.We know for every road how many minutes it takes to come from one city to another. 
    Mirko overheard in the car that one of the roads is under repairs, and that it is blocked, but didn't konw exactly which road. It is possible to come from Marica's city to Mirko's no matter which road is closed. 
    Marica will travel only by non-blocked roads, and she will travel by shortest route. Mirko wants to know how long will it take for her to get to his city in the worst case, so that he could make sure that his girlfriend is out of town for long enough.Write a program that helps Mirko in finding out what is the longest time in minutes it could take for Marica to come by shortest route by non-blocked roads to his city.

    InputEach case there are two numbers in the first row, N and M, separated by a single space, the number of towns,and the number of roads between the towns. 1 ≤ N ≤ 1000, 1 ≤ M ≤ N*(N-1)/2. The cities are markedwith numbers from 1 to N, Mirko is located in city 1, and Marica in city N. 
    In the next M lines are three numbers A, B and V, separated by commas. 1 ≤ A,B ≤ N, 1 ≤ V ≤ 1000.Those numbers mean that there is a two-way road between cities A and B, and that it is crossable in V minutes.OutputIn the first line of the output file write the maximum time in minutes, it could take Marica to come to Mirko.Sample Input

    5 6
    1 2 4
    1 3 3
    2 3 1
    2 4 4
    2 5 7
    4 5 1
    
    6 7
    1 2 1
    2 3 4
    3 4 4
    4 6 4
    1 5 5
    2 5 2
    5 6 5
    
    5 7
    1 2 8
    1 4 10
    2 3 9
    2 4 10
    2 5 1
    3 4 7
    3 5 10

    Sample Output

    11
    13
    27

    给定时间是5s,不会超时

    题意

    给出的所有路中存在一条路正在修建

    要从1走到n,找出最短路径的最大长度

    思路:

    先一遍dijkstra找到最短路径长度,并且记录路径

    再逐个删掉通向最短路径的每一条边换成别的能够到达终点的路

    计算其最大长度

    难点:在记录路径上

      1 #include<iostream>
      2 #include<string.h>
      3 #include<cmath>
      4 #include<iomanip>
      5 #define inf 0x3f3f3f3f
      6 #define ios() std::ios::sync_with_stdio(false)
      7 #define cin() cin.tie(0)
      8 #define cout() cout.tie(0)
      9 #define mem(a) memset(a,0,sizeof(a))
     10 using namespace std;
     11 
     12 int vertex[1001];//记录首先求得的最短路径所经过的顶点
     13 int n,m;//n顶点总个数,m是边数
     14 int e[1001][1001];
     15 int dis[1001],dis1[1001],dis2[1001];
     16 int book[1001];
     17 int flag;
     18 
     19 void init()
     20 {
     21     for(int i=1; i<=n; i++)
     22     {
     23         for(int j=1; j<=n; j++)
     24         {
     25             if(i==j)
     26                 e[i][j]=0;
     27             else
     28                 e[i][j]=inf;
     29             }
     30     }
     31 }
     32 
     33 int  dijkstra(int *dis)
     34 {
     35     mem(book);mem(dis1);mem(dis2);
     36     for(int i=1;i<=n;i++)
     37         dis[i]=e[1][i];
     38     book[1]=1;
     39     int u;
     40     for(int i=2;i<=n;i++)
     41     {
     42         int minn=inf;
     43         for(int j=1;j<=n;j++)
     44         {
     45             if(book[j]==0&&dis[j]<minn)
     46             {
     47                 u=j;
     48                 minn=dis[j];
     49             }
     50         }
     51         book[u]=1;
     52         for(int k=1;k<=n;k++)
     53         {
     54             if(e[u][k]<inf&&dis[u]+e[u][k]<dis[k])
     55             {
     56                 dis[k]=dis[u]+e[u][k];
     57                 if(flag)//如果flag=1,则开始记录最短路的路径
     58                 {
     59                     vertex[k]=u;//记录顶点
     60                 }
     61             }
     62         }
     63     }
     64     return dis[n];
     65 }
     66 
     67 int main()
     68 {
     69     ios();cin();cout();
     70     int x,y,z;
     71     while(cin>>n>>m)
     72     {
     73         mem(dis);mem(vertex);mem(e);
     74         init();
     75         for(int i=0; i<m; i++)
     76         {
     77             cin>>x>>y>>z;
     78             e[x][y]=z;
     79             e[y][x]=z;
     80         }
     81 
     82         flag=1;mem(dis1);
     83         fill(vertex,vertex+1001,1);
     84         int shortest=dijkstra(dis1);
     85       //  cout<<shortest<<"**"<<endl;
     86 
     87 //        for(int i=n;i>=2;i--)
     88 //            cout<<vertex[i]<<" ";
     89 
     90         flag=0;//通过控制flag决定是否记录路径
     91         int w=n;//一步步倒回去替换掉每一段的距离且变成无穷大,寻找其他边
     92         int ans=shortest;
     93         while(w!=1)//   通过while去控制是否已经倒退回去遍历到所有的边
     94         {
     95           //  cout<<"***"<<endl;
     96             int frontt=vertex[w];
     97             int ori=e[frontt][w];
     98             e[frontt][w]=e[w][frontt]=inf;//第一次进去先变最后一条边
     99             mem(dis2);
    100             int ww=dijkstra(dis2);
    101           //  if(ww!=inf)
    102                 ans=max(ww,ans);
    103             e[frontt][w]=e[w][frontt]=ori;
    104             w=frontt;
    105            // 变成无穷大之后再进行下一次找的时候可能会需要到这条边,所以需要恢复
    106         }
    107         cout<<ans<<endl;
    108     }
    109     return 0;
    110 }
    View Code
  • 相关阅读:
    Could not load file or assembly Microsoft.SqlServer.management.sdk.sfc version 11.0.0.0
    代码覆盖率 (Code Coverage)从简到繁 (一)
    vscode配置自动格式化eslint 配置模板
    无效的源发行版: 10
    java读取一个文件写入另外一个文件
    notepad++使用正则表达式匹配
    jsp页面返回字符串而非方法执行后取得的数据?
    maven的原始setting.xml文件,自带阿里云镜像,之前配的时候出错,保存一下,注意可以在localRepository处设置存储依赖的地址,大概在49到54行,我的是<localRepository>F:/MavenRepository</localRepository>,F盘要先有这个文件夹
    Java 8 lambda Stream list to Map key 重复 value合并到Collection
    BufferedReader.readLine()读取文件
  • 原文地址:https://www.cnblogs.com/OFSHK/p/11519270.html
Copyright © 2011-2022 走看看