Stockbroker Grapevine【POJ--1125】【Floyd】

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 suchtransmission is possible at all.

题意：一个股票经纪人在散播谣言，多组输入，每组输入n表示总共有n个人，接下来n行，每行输入一个m和m对数，第i行就表示第i个人有m个忠实者，那么m对数就表示第i个人对第m1个人传播谣言需要m2分钟，求从哪个人开始传播能让所有人知道这个谣言并用时最短，如果有，就输出其编号及传播时间；如果没有输出“disjoint”。

思路：用Floyd去更新任意两人之间传播谣言的所用时间，最后求出每条能通的路径的最大权的最小值且每两点都能通。

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

#include <cstdio>#include <cstring>#include <algorithm>#define INF 0x3f3f3f3fusing namespace std;int dj[150][150],n;void floyd(){    int k,tt,mm,flag;    for(int k=1; k<=n; k++)         //Floyd更新两人间传播谣言的所用时间    {        for(int i=1; i<=n; i++)        {            for(int j=1; j<=n; j++)            {                dj[i][j]=min(dj[i][j],dj[i][k]+dj[k][j]);            }        }    }    mm=INF;    for(int i=1; i<=n; i++)    {        flag=1;        tt=0;        for(int j=1; j<=n; j++)        {            if(i==j)continue;            if(dj[i][j]==INF)flag=0;//如果有两点不通，则进行标记说明不能从第i个人开始传播            tt=max(dj[i][j],tt);        }        if(flag&&tt<mm)        {            mm=tt;            k=i;        }    }    if(mm==INF)        printf("disjoint\n");    else        printf("%d %d\n",k,mm);}int main(){    //freopen("lalala.text","r",stdin);    int m,x,y;    while(~scanf("%d",&n))    {        if(n==0)            break;        memset(dj,INF,sizeof(dj));        for(int i=1; i<=n; i++)        {            scanf("%d",&m);            for(int j=0; j<m; j++)            {                scanf("%d %d",&x,&y);                dj[i][x]=y;            }        }        floyd();    }    return 0;}