ACM: 图论题 poj 1125 (题意读好久…

                                                     Stockbroker Grapevine

Description

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

Unfortunately for you, stockbrokers only trust information comingfrom their "Trusted sources" This means you have to take intoaccount the structure of their contacts when starting a rumour. Ittakes a certain amount of time for a specific stockbroker to passthe rumour on to each of his colleagues. Your task will be to writea program that tells you which stockbroker to choose as yourstarting point for the rumour, as well as the time it will take forthe rumour to spread throughout the stockbroker community. Thisduration is measured as the time needed for the last person toreceive 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 thenumber of people who they have contact with, who these people are,and the time taken for them to pass the message to each person. Theformat of each stockbroker line is as follows: The line starts withthe number of contacts (n), followed by n pairs of integers, onepair for each contact. Each pair lists first a number referring tothe contact (e.g. a '1' means person number one in the set),followed by the time in minutes taken to pass a message to thatperson. There are no special punctuation symbols or spacingrules.

Each person is numbered 1 through to the number of stockbrokers.The time taken to pass the message on will be between 1 and 10minutes (inclusive), and the number of contacts will range between0 and one less than the number of stockbrokers. The number ofstockbrokers will range from 1 to 100. The input is terminated by aset of stockbrokers containing 0 (zero) people.

Output

For each set of data, your program must output a single linecontaining the person who results in the fastest messagetransmission, and how long before the last person will receive anygiven message after you give it to this person, measured in integerminutes.
It is possible that your program will receive a network ofconnections that excludes some persons, i.e. some people may beunreachable. If your program detects such a broken network, simplyoutput the message "disjoint". Note that the time taken to pass themessage from person A to person B is not necessarily the same asthe time taken to pass it from B to A, if such transmission ispossible 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

题意: 股票经纪人,想要传播一个谣言,给每一个同事,但是每一个股票经纪人只相信自己那part人.
         要求求出选择哪个开始可以让谣言传播的最快.

个人思路:
               1.一开是想到是最小生成树问题.后面发现多次迭代的选择股票经纪人跟floyd思路类似.
               2.还是选择最短路径求解 floyd算法.
               3.最后要选择出最后一个人收到谣言的时间.(问题的转化: 谣言传播是多人同步进行.)

代码:

#include <cstdio>
#include <iostream>
#include <cstring>
using namespace std;
#define MAX 1000
const int INF = (1<<20);

int dist[MAX][MAX];
int n;
int point , maxsize , minsize;

void read_graph()
{
    int i ,j;
    for(i = 1; i<= n; ++i)
      for(j = 1; j <= n; ++j)
         dist[i][j] = (i == j ? 0 : INF);
         
    int k;
    int v ,w;
    for(i = 1; i<= n; ++i)
    {
      scanf("%d",&k);
      for(j = 1; j <= k; ++j)
      {
         scanf("%d%d",&v,&w);
         dist[i][v] = w;
      }
    }
}

void floyd()
{
    for(int k =1; k <= n; ++k)
    {
      for(int i = 1; i <= n; ++i)
      {
         for(int j = 1; j <= n; ++j)
         {
            if(dist[i][j] > dist[i][k] +dist[k][j])
            {
               dist[i][j] = dist[i][k] + dist[k][j];
            }
         }
      }
    }
}

void result()
{
    int i ,j;
    minsize =INF;
    for(i = 1; i<= n; ++i)
    {
      maxsize = 0;
      for(j = 1; j <= n; ++j)
      {
         if(i != j &&dist[i][j] > maxsize)
         {
            maxsize = dist[i][j];
         }
      }
      
      if(maxsize < minsize)
      {
         minsize = maxsize;
         point = i;
      }
    }
}

int main()
{
//   freopen("input.txt","r",stdin);
   while(scanf("%d",&n) != EOF&& n != 0)
    {
      read_graph();
      floyd();
      result();
      
      if(minsize == INF)
         printf("disjoint\n");
      else
         printf("%d %d\n",point,minsize);
    }
    
    return0;
}