zoukankan      html  css  js  c++  java
  • POJ 1125 Stockbroker Grapevine(任意两点的最短路,FLoyed 算法)

    Stockbroker Grapevine
    Time Limit: 1000MS   Memory Limit: 10000K
    Total Submissions: 21124   Accepted: 11474

    Description

    Stockbrokers are known to overreact to rumours. You have been contracted to develop a method of spreading disinformation amongst the stockbrokers to give your employer the tactical edge in the stock market. For maximum effect, you have to spread the rumours in the fastest possible way.

    Unfortunately for you, stockbrokers only trust information coming from their "Trusted sources" This means you have to take into account the structure of their contacts when starting a rumour. It takes a certain amount of time for a specific stockbroker to pass the rumour on to each of his colleagues. Your task will be to write a program that tells you which stockbroker to choose as your starting point for the rumour, as well as the time it will take for the rumour to spread throughout the stockbroker community. This duration is measured as the time needed for the last person to receive the information.

    Input

    Your program will input data for different sets of stockbrokers. Each set starts with a line with the number of stockbrokers. Following this is a line for each stockbroker which contains the number of people who they have contact with, who these people are, and the time taken for them to pass the message to each person. The format of each stockbroker line is as follows: The line starts with the number of contacts (n), followed by n pairs of integers, one pair for each contact. Each pair lists first a number referring to the contact (e.g. a '1' means person number one in the set), followed by the time in minutes taken to pass a message to that person. There are no special punctuation symbols or spacing rules.

    Each person is numbered 1 through to the number of stockbrokers. The time taken to pass the message on will be between 1 and 10 minutes (inclusive), and the number of contacts will range between 0 and one less than the number of stockbrokers. The number of stockbrokers will range from 1 to 100. The input is terminated by a set of stockbrokers containing 0 (zero) people.

    Output

    For each set of data, your program must output a single line containing the person who results in the fastest message transmission, and how long before the last person will receive any given message after you give it to this person, measured in integer minutes.
    It is possible that your program will receive a network of connections that excludes some persons, i.e. some people may be unreachable. If your program detects such a broken network, simply output the message "disjoint". Note that the time taken to pass the message from person A to person B is not necessarily the same as the time taken to pass it from B to A, if such transmission is possible at all.

    Sample Input

    3
    2 2 4 3 5
    2 1 2 3 6
    2 1 2 2 2
    5
    3 4 4 2 8 5 3
    1 5 8
    4 1 6 4 10 2 7 5 2
    0
    2 2 5 1 5
    0

    Sample Output

    3 2
    3 10

    Source

     
     
    /*
    POJ 1125
    G++ 0ms 172K
    */
    
    #include<stdio.h>
    #include<iostream>
    #include<algorithm>
    using namespace std;
    const int MAXN=110;
    const int INF=0x3f3f3f3f;
    
    void floyed(int dis[][MAXN],int n)//结点编号1-n
    {
        int i,j,k;
        for(k=1;k<=n;k++)
           for(i=1;i<=n;i++)
             for(j=1;j<=n;j++)
               if(dis[i][j]>dis[i][k]+dis[k][j])
                  dis[i][j]=dis[i][k]+dis[k][j];
    }
    int dis[MAXN][MAXN];
    int main()
    {
        int n;
        int m;
        int a,b;
        while(scanf("%d",&n),n)
        {
            for(int i=1;i<=n;i++)
              for(int j=1;j<=n;j++)
              {
                  if(i==j)dis[i][j]=0;
                  else dis[i][j]=INF;
              }
            for(int i=1;i<=n;i++)
            {
                scanf("%d",&m);
                while(m--)
                {
                    scanf("%d%d",&a,&b);
                    dis[i][a]=min(dis[i][a],b);
                }
            }
            floyed(dis,n);
            int ans=INF;
            int t;
            int temp;
            int j;
            for(int i=1;i<=n;i++)
            {
                temp=0;
                for(j=1;j<=n;j++)
                  if(j!=i&&dis[i][j]>temp)
                    temp=dis[i][j];
                if(temp<ans)
                {
                    ans=temp;
                    t=i;
                }
            }
            printf("%d %d\n",t,ans);
        }
        return 0;
    }
  • 相关阅读:
    Spring-MVC
    Spring与MyBatis整合(哇,快来看,他写的真好,简单易懂哎)
    Spring事务终极版...
    JdbcTemplate(增删改查以及以注释实现增删改查)
    AOP进阶
    MyBatis增删改查
    JSP的数据交互
    JSP数据交互
    JSP中9大内置对象
    动态网页开发基础
  • 原文地址:https://www.cnblogs.com/kuangbin/p/2645714.html
Copyright © 2011-2022 走看看