OpenJudge/Poj 1125 Stockbroker Grapevine

1.链接地址：

http://poj.org/problem?id=1125

http://bailian.openjudge.cn/practice/1125

2.题目：

Stockbroker Grapevine

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 24810		Accepted: 13674

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

3.思路：

4.代码：

 #include <iostream>
 #include <cstdio>

 using namespace std;
 int dis[][];//最多为100，为了防止溢出订为105
 void Floyd(int n)
 {
  int i,j,k;
  for(i=;i<=n;i++)
   for(j=;j<=n;j++)
    for(k=;k<=n;k++)
        //dp，使用弗洛伊德算法
     if(dis[j][i]+dis[i][k]<dis[j][k]){
      dis[j][k]=dis[j][i]+dis[i][k];
     }
 }
 void Min(int n)
 {
  int node,min=0xfffff;//0xfffff为最大int数
  int i,j;
  for(i=;i<=n;i++){
   int max=;
   for(j=;j<=n;j++){
    if(i==j) continue;
    //找出个人传到每个人的最大的时间
    if(dis[i][j]>max) max=dis[i][j];
   }
   if(max<min){
       //比较每个人的最大时间，寻找最小的一人
    min=max;
    node=i;
   }
  }
  if(min<0xffff){
   printf("%d %d\n",node,min);
  }
  else{
   printf("disjoint\n");
  }

 }
 int main()
 {
  int n,i,j,k,m;

  //初始化邻接矩阵
  while(scanf("%d",&n)&&n){
   for(i=;i<=n;i++){
    for(j=;j<=n;j++){
     dis[i][j]=0xfffff;
    }
   }
   //m为人数
   for(i=;i<=n;i++){
    scanf("%d",&m);

    for(j=;j<m;j++){//k为认识的人
     scanf("%d",&k);
     scanf("%d",&dis[i][k]);
    }
   }
   Floyd(n);
   Min(n);
  }
  return ;
 }