Stockbroker Grapevine
Time Limit: 1000MS
Memory Limit: 10000K
Total Submissions: 27977
Accepted: 15527
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
Source
Southern African 2001
题目大意：有N个股票经济人，他们之间可以传递信息，但是他们只相信他们认为可靠的人的信息。
现在由某个人开始传信息，怎么能在最短的时间内让所有人都接收到消息。这个时间取决于最后一
个人收到信息的时间。如果没有一个人能使所有人都接收到信息，则输出"disjoint"，否则，就输出
最短的时间和这个人的编号。
思路：可以看做是N个点，M条单向边。建立一个图，然后用Floyd求多源最短路径。之后，遍历所
有的结点，找到符合要求的那个人编号。不存在就输出"disjoint"。
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
using namespace std;
const int MAXN = 110;
const int INF = 0xffffff0;
int Map[MAXN][MAXN], Dist[MAXN][MAXN];
void Floyd(int N)
{
for(int i = 1; i <= N; ++i)
{
for(int j = 1; j <= N; ++j)
Dist[i][j] = Map[i][j];
}
for(int k = 1; k <= N; ++k)
{
for(int i = 1; i <= N; ++i)
{
for(int j = 1; j <= N; ++j)
{
if(Dist[i][k] != INF && Dist[k][j] != INF && Dist[i][k] + Dist[k][j] < Dist[i][j])
Dist[i][j] = Dist[i][k] + Dist[k][j];
}
}
}
}
int main()
{
int N;
while(~scanf("%d",&N) && N)
{
for(int i = 1; i <= N; ++i)
for(int j = 1; j <= N; ++j)
Map[i][j] = INF;
for(int i = 1; i <= N; ++i)
{
int d;
scanf("%d",&d);
while(d--)
{
int to,dist;
scanf("%d%d",&to,&dist);
Map[i][to] = dist;
}
}
Floyd(N);
int ans = INF,pos,tMin;
for(int i = 1; i <= N; ++i)
{
int flag = 0;
tMin = 0;
for(int j = 1; j <= N; ++j)
{
if(i != j && Dist[i][j] == INF)
{
flag = 1;
break;
}
if(i != j && Dist[i][j] > tMin)
{
tMin = Dist[i][j];
}
}
if(flag == 0 && tMin < ans)
{
ans = tMin;
pos = i;
}
}
if(ans == INF)
printf("disjoint\n");
else
printf("%d %d\n",pos,ans);
}
return 0;
}