POJ 1125 Stockbroker Grapevine

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
Southern African 2001

 

题目大意：

第一行为证券经纪人的数量n。接下来的n行给出每位证券经纪人能够联系上的其他经纪人，以及联系他们所要花费的时间。需要确定将消息首先传递给哪一位经纪人，能使消息传遍所有经纪人所花费的时间最小。注意，证券经纪人可以同时将消息传递给其他人，时间以单独传递时的最长时间计算。

 

我的思路：

这是一个“最短路径”的问题，典型求解方法有Dijkstra算法和Floyd算法。Floyd算法能够有效地求解每一对顶点之间的最短路径，我认为能更有效地解决本题所述问题。Floyd算法的迭代公式如下：

A₀[i][j]=Cost[i][j]

A_k+1[i][j]=min{A_k[i][j],A_k[i][k+1]+A_k[k+1][j]}

在本题中，首先需要根据输入数据建立联系关系图的邻接矩阵，然后根据Floyd算法求的每个顶点之间的最短路径矩阵。之后，对最短路径矩阵进行搜索——求取每一行的最大值中的最小值。所得结果即为最优解。

 

我的代码：

#include<iostream>
#include<fstream>

using namespace std;

#define ROW 101
#define COL 101
#define MAX  9999

int main()
{
 ifstream cin("data.txt");
 int cost[ROW][COL];

 int i,j,k;
 int person;                   // 人数
 int number;                   // 某人能联系上的人数
 int index;                    // 能够联系上的人的序号

 while( cin>>person &&person>0)
 {
  memset( cost,MAX,sizeof(cost) );

  for( i=0; i<person; ++i )
  {
   cost[i][i]=0;                  // 连接矩阵对角线上元素等于0
   cin>>number;
   for( j=0; j<number; ++j )
   {
    cin>>index;
    cin>>cost[i][index-1];     // 输入第i人联系第index人耗时
   }
  }

  // floyd 算法：迭代过程
  for( k=0; k<person; ++k )
   for( i=0; i<person; ++i)
    for( j=0; j<person; ++j )
    {
     if( cost[i][j]>cost[i][k]+cost[k][j] )
     {
      cost[i][j]=cost[i][k]+cost[k][j];
     }
    }
  // 搜索消息传播第一人，和用时
  int min = MAX;
  int id = 1;
  for( i=0; i<person; ++i )
  {
   int temp=cost[i][0];
   for( j=0; j<person; ++j )
   {
    if( cost[i][j]>temp )
    {
     temp = cost[i][j];
    }
   }

   if( temp<min )
   {
    min = temp;
    id = i+1;
   }
  }

  cout<<id<<" "<<min<<endl;
 }
 return 0;
}