zoukankan      html  css  js  c++  java
  • POJ1125 Stockbroker Grapevine

     
    Time Limit: 1000MS   Memory Limit: 10000K
    Total Submissions: 34374   Accepted: 19027

    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

     
    看题时间大于做题时间系列。
    求每个点到其他点的最短路,然后枚举从每个点出发,找最优解即可。
     
     1 /**/
     2 #include<iostream>
     3 #include<cstdio>
     4 #include<cmath>
     5 #include<cstring>
     6 #include<algorithm>
     7 using namespace std;
     8 const int INF=0x3f3f3f3f;
     9 const int mxn=200;
    10 int mp[mxn][mxn];
    11 int ans=INF,pos=0;
    12 int n,m;
    13 void read(){
    14     int x,t;
    15     int i,j;
    16     for(i=1;i<=n;i++){
    17         scanf("%d",&m);
    18         for(j=1;j<=m;j++){
    19             scanf("%d%d",&x,&t);
    20             mp[i][x]=t;
    21         }
    22     }
    23     return;
    24 }
    25 void floyd(){
    26     int i,j,k;
    27     for(k=1;k<=n;k++)
    28      for(i=1;i<=n;i++)
    29       for(j=1;j<=n;j++){
    30           mp[i][j]=min(mp[i][j],mp[i][k]+mp[k][j]);
    31       }
    32     return;
    33 }
    34 int main(){
    35     while(scanf("%d",&n) && n){
    36         memset(mp,0x3f,sizeof mp);
    37         ans=INF;
    38         pos=0;
    39         read();
    40         floyd();
    41         int i,j;
    42         for(i=1;i<=n;i++){
    43             int mx=0;
    44             for(j=1;j<=n;j++){
    45                 if(j==i)continue;
    46                 mx=max(mx,mp[i][j]);
    47             }
    48             if(ans>mx){
    49                 ans=mx;
    50                 pos=i;
    51             }
    52         }
    53         if(ans!=INF)printf("%d %d
    ",pos,ans);
    54         else printf("disjoint
    ");
    55     }
    56     return 0;
    57 }
  • 相关阅读:
    PHP unicode与普通字符串的相互转化
    PHP 日期之间所有日期
    PHP Excel导入日期单元格处理
    JS base64文件转化blob文件
    Mvc 刷新PartialView
    WebGL绘制变幻光斑
    WebGL笔记(四):初步封装
    WebGL笔记(目录)
    [JavaScript/canvas] 创建基于坐标访问的图形数据对象
    WebGL笔记(五):封装顶点和颜色等数组数据(二)
  • 原文地址:https://www.cnblogs.com/SilverNebula/p/5837736.html
Copyright © 2011-2022 走看看