Stockbroker Grapevine（最短路）

  　　　　　　　　　　　　　　　　　　　　　　　　　　　poj——1125 Stockbroker Grapevine

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 36112		Accepted: 20033

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

题目翻译：

经纪人称过度的传言。你已经签约开发中传播假情报给雇主股票在股票市场上的战术优势的方法。为了达到最大的效果，你必须以最快的速度传播谣言。
不幸的是你，股票经纪人信息只信任来自他们的“可信任的来源”，这意味着你要考虑到他们的接触结构时的谣言。一个特定的股票经纪人把谣言传给他的每一个同事都需要一定的时间。你的任务是写一个程序，告诉你选择哪个股票经纪人作为你的出发点的谣言，以及时间将需要在整个股票经纪人社区传播谣言。此持续时间被测量为最后一个人接收信息所需的时间。
输入
你的程序将输入的数据不同的股票经纪人。每一组从一个股票经纪人的人数。以下是每个股票经纪人的一行，其中包括他们接触过的人，这些人是谁，以及他们把信息传递给每个人的时间。每个股票经纪人行的格式如下：该行开始与接触数（n），其次是n对整数，一对对每个接触。每对先列出一个数字，指的是联系人（如“1”指的是集合中的一个人），接着是将信息传递给那个人的分钟。没有特殊的标点符号或间距规则。
每个人的编号为1到一些股票经纪人。来传递信息的时间将是1至10分钟（含），并将0个触点数小于股票的数量之间的距离。股票的数量范围从1到100。输入是一套股票经纪人含有0（零）的人。
输出
对于每一组数据，你的程序必须输出一个包含最快消息传输结果的人的一行，以及在最后一个人收到该信息后，要多长时间才能接收到任何给定的信息。
这是可能的，您的程序将收到一个网络的连接，排除一些人，即一些人可能遥不可及。如果您的程序检测到这样一个断开的网络，只需输出消息“不相交”。请注意，从一个人到另一个人传递消息所花费的时间不一定与把它从B传给A的时间相同，如果这样的传输是完全可能的。

思路：

这道题，我们只需要先使用floyd求出任意两点间的距离。

之后，我们可以去枚举我们选取的最初传递人，然后求出他传递给所有人所需的时间，再在所有这些枚举的时间里面找到一个最小值就可以了。

代码：

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#define N 101
#define maxn 9999999
using namespace std;
int n,m,x,y,z,dis[N][N],tot,ans,p;
struct Edge
{
    int x,y,z;
}edge[N*N];
int read()
{
    ,f=;
    char ch=getchar();
    ')
    {
        ;
        ch=getchar();
    }
    ')
    {
        x=x*+ch-';
        ch=getchar();
    }
    return x*f;
}
void begin(int n,int m)
{
    ;i<=n;i++)
     ;j<=m;j++)
      dis[i][j]=maxn*(i!=j);
}
int main()
{
    while(n=read(),n)
    {
        tot=;
        begin(N,N);
        ;i<=n;i++)
        {
            m=read();
            ;j<=m;j++)
             {
                 x=read(),y=read();
                 dis[i][x]=y;
             }
        }
        ;k<=n;k++)
         ;i<=n;i++)
          ;j<=n;j++)
           dis[i][j]=min(dis[i][j],dis[i][k]+dis[k][j]);
        int maxx;
        bool b=true;
        ans=maxn;
        ;i<=n;i++)
        {
            maxx=;
            ;j<=n;j++)
            {
                if(i==j) continue;
                if(maxx<dis[i][j]) maxx=dis[i][j];
            }
            if(ans>maxx) ans=maxx,p=i;
            if(maxx!=maxn) b=false;
        }
        if(b) printf("disjoint\n");
        else printf("%d %d\n",p,ans);
    }
    ;
}