POJ - 1125 Stockbroker Grapevine（Java）

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

超简易题意：
输出一名满足下列条件的经纪人编号：

1、能联系到其他所有经纪人

2、联系速度最快（最短路）

再输出这位经纪人联系最慢那一位的编号。

因为要求任意两点间的最短路，所以用佛洛依德算法，一口气生成所有最短路，然后按题意筛选。具体看代码注释吧。
ac代码：

public class Main {
    static int[][] dis = new int[105][105];
    static final int INF = 100000;

    public static void main(String[] args) {
        Scanner reader = new Scanner(System.in);
        int n = reader.nextInt();
        while (n != 0) {
            for (int i = 0; i < 100; i++) {
                Arrays.fill(dis[i], INF);
            }
            for (int i = 0; i < n; i++) {
                int know = reader.nextInt();
                for (int j = 0; j < know; j++) {
                    dis[i][reader.nextInt() - 1] = reader.nextInt();
                }
            }
            System.out.println(floyd(n));
            n = reader.nextInt();
        }
    }

    public static String floyd(int n) {
        for (int k = 0; k < n; k++) {
            for (int i = 0; i < n; i++) {
                for (int j = 0; j < n; j++) {
                    boolean cantGoThrough = i == j || dis[i][k] == INF || dis[k][j] == INF;
                    if (cantGoThrough) {
                        continue;
                    }
                    dis[i][j] = Math.min(dis[i][j], dis[i][k] + dis[k][j]);
                }
            }
        }
        // 开始筛选
        int key = 0, ans = INF;
        for (int i = 0; i < n; i++) {
            int max = 0;
            for (int j = 0; j < n; j++) {
                if (i != j && dis[i][j] > max) {// 如果不能传播到所有人，自然会记录到一个INF，自动无视
                    max = dis[i][j];
                }
            }
            // System.out.println(max);
            if (max < ans) {// 最大值都小了，其他的不用比了，选最大值最小那个记录下来
                ans = max;
                key = i;
            }
        }
        return ans == INF ? "disjoint" : (key + 1) + " " + ans;
    }
}