Stockbroker Grapevine （弗洛伊德算法）

Stockbroker Grapevine

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 20000/10000K (Java/Other)

Total Submission(s) : 28   Accepted Submission(s) : 16

Problem 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.<br><br>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.<br><br>

 

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.<br>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

32 2 4 3 52 1 2 3 62 1 2 2 253 4 4 2 8 5 31 5 84 1 6 4 10 2 7 5 202 2 5 1 50

 

Sample Output

3 23 10

 

Source

PKU

题意：

股票传播谣言它需要特定股票经纪人和一定的时间把谣言传递给他的每一位同事。告诉您选择哪一个股票经纪人作为谣言的出发点和所花费多少时间将谣言扩散到整个社会的股票经纪人，  这一期限是衡量过去的人收到信息所需的时间。你的程序必须输出一行，包括的信息有传输速度最快的人，以及在最后一个人收到消息后，所总共使用的时间（整数分钟计算）。  你的程序可能会收到的一些关系会排除一些人，也就是有些人可能无法访问。如果你的程序检测到这样一个破碎的网络，  只需输出消息“disjoint”。请注意，所花费的时间是从A传递消息到B，B传递信息到A不一定是花费同样的传递时间，但此类传播也是可能的。 、

思路：

简单的弗洛伊德变形即可！

代码：

#include <iostream>#include <cstring>#include <string>#include <stdio.h>#define maxn 0x3f3f3f3fusing namespace std;int mp[205][205];int main(){    int n,m,i,j,k,t,max,min;    while(cin>>n,n)    {      /* for(i=1;i<=n;i++){            mp[i][i]=0;            for(j=1;j<i;j++)                mp[i][j]=mp[j][i]=maxn;        }*/        memset(mp,maxn,sizeof(mp));        for(i=1;i<=n;i++)        {            cin>>m;            while(m--){cin>>k>>t;mp[i][k]=t;}        }        for(k=1;k<=n;k++)            for(i=1;i<=n;i++)            for(j=1;j<=n;j++)        {            if(i!=k&&j!=k&&i!=j&&mp[i][j]>mp[i][k]+mp[k][j]) mp[i][j]=mp[i][k]+mp[k][j];        }        min=maxn;        max=0;        for(i=1;i<=n;i++)        {            max=0;            for(j=1;j<=n;j++)            {                if(i==j)continue;                if(mp[i][j]>max) max=mp[i][j];            }            if(max<min)            {                min=max;                t=i;            }        }        if(min==maxn)             cout<<"disjoint"<<endl;        else        cout<<t<<' '<<min<<endl;    }    return 0;}