ACM学习-POJ-1125-Stockbroker Grapevine

ゝ一世哀愁。 2022-09-08 05:57 166阅读 0赞

**菜鸟学习ACM，纪录自己成长过程中的点滴。**

**学习的路上，与君共勉。**

**ACM学习-POJ-1125-Stockbroker Grapevine**

Stockbroker Grapevine

<table> 
 <tbody> 
  <tr> 
   <td><strong>Time Limit:</strong>&nbsp;1000MS</td> 
   <td>&nbsp;</td> 
   <td><strong>Memory Limit:</strong>&nbsp;10000K</td> 
  </tr> 
  <tr> 
   <td><strong>Total Submissions:</strong>&nbsp;24008</td> 
   <td>&nbsp;</td> 
   <td><strong>Accepted:</strong>&nbsp;13205</td> 
  </tr> 
 </tbody> 
</table>

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][]

**题目要求：**

描述   
众所周知，证券经纪业依靠的就是过度的传言。您需要想出股票经纪人中传播假情报的方法，让您的雇主在股票市场的占据优势。为了获得最大的效果，你必须蔓延最快的方式谣言。   
不幸的是你，股票经纪人信息只信任他们的“可靠来源”，这意味着你在你传播谣言之前必须考虑到他们的接触结构。它需要特定股票经纪人和一定的时间把谣言传递给他的每一位同事。你的任务将是写一个程序，告诉您选择哪一个股票经纪人作为谣言的出发点和所花费多少时间将谣言扩散到整个社会的股票经纪人。这一期限是衡量过去的人收到信息所需的时间。   
输入   
你的程序包含多组股票经纪人的输入数据。每组以股票经纪人的人数开始。接下来的几行是每个经纪人与其他人接触的一些信息，包括这些人都是谁，以及将讯息传达到他们所需的时间。每个经纪人与其他人接触信息的格式如下：开头的第一个数表示共有n个联系人，接下来就有n对整数。每对整数列出的第一个数字指的是一个联系人（例如，一个'1'是指编号1的人），其次是在传递一个信息给那个人时所采取分钟的时间。没有特殊的标点符号或空格规则。   
每个人的编号为1至经纪人数目。所花费的传递时间是从1到10分钟（含10分种）。股票经纪的人数范围是从1到100。当输入股票经纪人的人数为0时，程序终止。   
输出   
在对于每一组数据，你的程序必须输出一行，包括的信息有传输速度最快的人，以及在最后一个人收到消息后，所总共使用的时间（整数分钟计算）。   
你的程序可能会收到的一些关系会排除一些人，也就是有些人可能无法访问。如果你的程序检测到这样一个破碎的网络，只需输出消息“disjoint”。请注意，所花费的时间是从A传递消息到B，B传递信息到A不一定是花费同样的传递时间，但此类传播也是可能的。

**题目分析：**

**类似于求A,B两点的最短距离。 让我一下想到了Floyd算法。**

**（不懂这个算法的同学可以看看我的另外一篇博客 http://blog.csdn.net/hitwhylz/article/details/11990069）**

**其他的也没什么特别的。 直接上AC代码，不过这次是C++。 应该也不影响阅读。**

#include <stdio.h>
    
    const int N = 105;
    const int MAX = 0xfffffff;
    
    int map_arr[N][N];
    int n;
    
    void floyd()
    {
        for (int k = 0; k < n; ++k)
        {
            for (int i = 0; i < n; ++i)
            {
                for (int j = 0; j < n; ++j)
                {
                    if (map_arr[i][j] > map_arr[i][k] + map_arr[k][j])
                    {
                        map_arr[i][j] = map_arr[i][k] + map_arr[k][j];
                    }
                }
            }
        }
    }
    
    int main()
    {
        int m, mate, dis;
        int i, j;
        int disjoint;
        int ans, tmin, pnt;
    
        while (~scanf("%d", &n))
        {
            for (i = 0; i < n; ++i)
            {
                for (int j = 0; j < n; ++j)
                {
                    if (i != j)
                        map_arr[i][j] = MAX;
                    else
                        map_arr[i][j] = 0;
                }
            }
            for (i = 0; i < n; ++i)
            {
                scanf("%d", &m);
                for (j = 0; j < m; ++j)
                {
                    scanf("%d%d", &mate, &dis);
                    --mate;
                    map_arr[i][mate] = dis;
                }
            }
            floyd();
            ans = MAX;
            for (int i = 0; i < n; ++i)
            {
                tmin = -1;
                disjoint = 0;
                for (int j = 0; j < n && !disjoint; ++j)
                {
                    if (i != j && map_arr[i][j] == MAX)
                        disjoint = 1;
                    if (i != j && map_arr[i][j] > tmin)
                        tmin = map_arr[i][j];
                }
                if (!disjoint && tmin < ans)
                {
                    disjoint = 0;
                    ans = tmin;
                    pnt = i;
                }
            }
            if (ans == MAX)
                printf("disjoint\n");
            else
                printf("%d %d\n", pnt + 1, ans);
        }
        return 0;
    }

[Southern African 2001]: http://poj.org/searchproblem?field=source&key=Southern+African+2001