POJ 1125：Stockbroker Grapevine
Stockbroker Grapevine
Time Limit: 1000MS
Memory Limit: 10000KB
64bit IO Format: %I64d & %I64u
Submit Status
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
題目給出了一個股票經紀人傳信息的網絡，第一個N代表這個網絡中有多少個股票經紀人。之後給出了每個股票經紀人的情況。他能夠傳給誰以及其時間，求誰傳達整個網絡的時間最短，最短時間又是多少。如果這個網絡本身是不聯通的，那就輸出disjoint。
發現這些圖論的算法不知道的時候特別神秘，然後知道每一個是干什麼的之後才發現很多都是用一個模板去做題，當然目前自己做的題目都是圖論當中比較簡單的，所以自己覺得容易，以後應用的時候要好好思考。
但就這個題目來說，直接floyd套用就好了。而且這道題的數據也很水。
代碼：
#include
#include
#include
#include
#include
#include
#pragma warning(disable:4996)
using namespace std;
int num;
int dis[105][105];
int dis_max[105];
void init()
{
int i,j;
for(i=1;i<=num;i++)
{
for(j=1;j<=num;j++)
{
if(i==j)
{
dis[i][j]=0;
}
else
{
dis[i][j]=1005;
}
}
}
}
int main()
{
int i,j,k,i_num;
while(cin>>num)
{
if(num==0)
break;
init();
for(i=1;i<=num;i++)
{
cin>>i_num;
int x,x_dis;
for(j=1;j<=i_num;j++)
{
cin>>x>>x_dis;
dis[i][x]=x_dis;
}
}
for(k=1;k<=num;k++)
{
for(i=1;i<=num;i++)
{
for(j=1;j<=num;j++)
{
if(dis[i][k]+dis[k][j]
