POJ 1125 Stockbroker Grapevine

时间：2015-07-25 19:48:23 阅读：281 评论：0 收藏：0 [点我收藏+]

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

题意：It takes a certain amount of time for a specific stockbroker to pass the rumour on to each of his colleagues.（这句话是重点·）

让你找一个人做为传播流言的源点，这个人传播给其他所有人的时间要是最短的，那么只需要先找到每个人传播给其他人其中所用时间最长的（同时进行传播，即只需要找到传播的最慢的那个时间），当每个人都当过源点遍历完后比较哪个人传播给其他人最快

#include<stdio.h>
#define INF 0x3f3f3f3f
#define N 110
#define min(a, b) (a < b ? a : b)
int G[N][N], n;
void Init()
{
    int i, j;
    for (i = 0; i <= n; i++)
    {
        for (j = 0; j <= n; j++)
        {
            if (i == j) G[i][j] = 0;
            else G[i][j] = INF;
        }
    }
}
void Dist()
{
    int i, j, k;
    for (k = 1; k <= n; k++)
    {
        for (i = 1; i <= n; i++)
        {
            for (j = 1; j <= n; j++)
                G[i][j] = min(G[i][j], G[i][k]+G[k][j]);
        }
    }
}
int main ()
{
    int m, a, b, i, j, ans, idex, Min;
    while (scanf("%d", &n), n)
    {
        Init();
        for (i = 1; i <= n; i++)
        {
            scanf("%d", &m);
            while (m--)
            {
                scanf("%d %d", &a, &b);
                G[i][a] = b;
            }
        }
        Dist();
        ans = INF;
        for (i = 1; i <= n; i++)
        {
            Min = 0;
            for (j = 1; j <= n; j++)
            {
                if (G[i][j] > Min)
                    Min = G[i][j];
            }
            if (Min < ans)
            {
                ans = Min;
                idex = i;
            }
        }
        if (ans == INF) printf("disjoint\n");
        else printf("%d %d\n", idex, ans);
    }
    return 0;
}

POJ 1125 Stockbroker Grapevine

原文：http://www.cnblogs.com/syhandll/p/4676313.html

踩

(0)

评论一句话评论（0）

分享档案

更多>

2021年09月23日 (328)
2021年09月24日 (313)
2021年09月17日 (191)
2021年09月15日 (369)
2021年09月16日 (411)
2021年09月13日 (439)
2021年09月11日 (398)
2021年09月12日 (393)
2021年09月10日 (160)
2021年09月08日 (222)