问题描述

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.

输入

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.

输出

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.

样例

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

题意

有n个股票经纪人，每当有谣言时来每个股票经纪人都在某一个时刻传给他m个的联系人。
当现在来了一个谣言的时候，哪个股票经纪人最先知道这个谣言的时候，能最快地把这个谣言散步到大家都知道，最短时间是多少。
如果没有哪个股票经纪人得知谣言后能散布给所有人，则输出“disjoint”。

题解

用Floyd求出任意两点间的最短路。
然后枚举每个股票经纪人散布给其他人的最大的时间。
如果这个时间最大值为INF,说明哪个股票经纪人都无法把谣言散布给其他人，则输出“disjoint”。
否则就是谣言散布到所有人身上是最短的时间。

#include<iostream> 
#include<algorithm>
using namespace std;
const int INF = 1005;
const int maxn = 1e2+5;
int dp[maxn][maxn]; 
int n,m;
void Floyd()
{
    for(int k=1;k<=n;++k)
    {
        for(int i=1;i<=n;++i)
        {
            for(int j=1;j<=n;++j)
            {
//                if(i!=j)
                dp[i][j] = min(dp[i][j],dp[i][k]+dp[k][j]);
            }
        }
    }
}
int main()
{
    int v,e;
    while(cin>>n)
    {
        if(n==0) break;
        
        for(int i=1;i<=n;++i)
        {
            for(int j=1;j<=n;++j)
            {
                dp[i][j]=INF;
            }
            dp[i][i]=0;    
        }
        
        for(int i=1;i<=n;++i)
        {
            cin>>m;
            for(int j=0;j<m;++j)
            {
                cin>>v>>e;
                dp[i][v]=e;            //顶点i到定点v之间有一条边，边权为e
            }
        }    
        
        Floyd();
        
//        for(int i=1;i<=n;++i)
//        {
//            for(int j=1;j<=n;++j)
//            {
//                cout<<dp[i][j]<<" ";
//            }
//            cout<<endl;
//        }
//        cout<<endl;
        
        int res = INF;
        int pos = 0;
        for(int i=1;i<=n;++i)
        {
            int max_e = -1;
            for(int j=1;j<=n;++j)
            {
                if(i==j) continue;
                if( dp[i][j] > max_e)
                {
                    max_e = dp[i][j];
                }
            }
            if(max_e < res)
            {
                res = max_e;
                pos = i;    
            }
        }
        
        if(res ==  INF)
        {
            cout<<"disjoint"<<endl;
        }else 
        {
            cout<<pos<<" "<<res<<endl;
        }
    }
    return 0;
}