POJ 1125 Stockbroker Grapevine

Time Limit:1000MS    Memory Limit:10000KB     64bit IO Format:%I64d & %I64u

Submit Status Practice POJ 1125

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

#include<stdio.h>
#define INF 0x3fffff
int map[1005][1005];
int n;
int ans[1005];
void floyed()
{
	int i,j,k,max,min,l;
	for(k=1;k<=n;k++)
	{
		for(i=1;i<=n;i++)
		{
			for(j=1;j<=n;j++)
			{
				if(map[i][j]>map[i][k]+map[k][j])
					map[i][j]=map[i][k]+map[k][j];
			}
		}
	}
    for(i=1;i<=n;i++)
	{
		max=map[i][1];
		for(j=1;j<=n;j++)
		{
			if(map[i][j]>max)
			max=map[i][j];
		}
		ans[i]=max;
	}
    min=INF;
	for(i=1;i<=n;i++)
	{
		if(min>ans[i])
		min=ans[i];
	}
	for(i=1;i<=n;i++)
	if(min==ans[i])
	{
		l=i;
		break;
	}
	printf("%d %d\n",i,min);
}
int main()
{
	int i,k,j,x,y;
	while(scanf("%d",&n),n)
	{
	  for(i=1;i<=n;i++)
	  {
		  for(j=1;j<=n;j++)
		  {
			  if(i==j)
			  map[i][j]=0;
			  else
			  map[i][j]=INF;
		  }
	  }
	  for(i=1;i<=n;i++)
	  {
		  scanf("%d",&k);
		  for(j=1;j<=k;j++)
		  {
			  scanf("%d%d",&x,&y);
			  map[i][x]=y;
		  }
	  }
	  floyed();
	}
	  return 0;
}
测试数据:
Sample Input

3                        //表示有多少个经济人
2 2 4 3 5                //2表示第一个经济人有多少个联系人，并且与联系人2联系需多长时间4，同理3,5
2 1 2 3 6                //2表示第二个经济人有多少个联系人，并且与联系人1联系需多长时间2，同理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                       //输出那个联系人3联系用时最少，并输出最少用时2
3 10
题目大致题意为联系人之间互相联系，且需要时间，求出联系人之间联系所有人后用时最短的那个联系人，还有在所有联系人中所用时间最少
先求出联系人与其余联系人之间联系用时最多，表示联系这些人需要的时间，然后把多有联系人联系其余联系人需要的时间作对比，求出用时
最少的联系人和时间，因为经济人的数量范围是1~100，所以在不超时(10^9)的情况下使用floyd比较简便,即100的3次方循环