POJ 1125 Stockbroker Grapevine 解题报告

分类：图论，最短路，Floyd算法

作者：ACShiryu

时间：2011-7-28

地址： ACShiryu's Blog

Stockbroker Grapevine

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 17227		Accepted: 9306

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

题目大意是股票经纪人要在一群人中散布一个谣言，而谣言只能在亲密的人中传递，题目各处了人与人之间的关系及传递谣言所用的时间，要求程序给出应以那个人为起点，可以在最短的时间内让所有的人都得知这个谣言。要注意从a到b传递的时间不一定等于从b到a的时间，如果没有方案能够让每一个人都知道谣言，则输出"disjoint"。（有关图的连通性，你懂得！但好像不用考虑这种情况一样能AC，只能说测试数据有点小水！）

题目数据的输入第一行为n，代表总人数，当n=0时结束程序，接着n行，第i+1行的第一个是一个整数t，表示第i个人与t个人的关系要好，接着有t对整数，每对的第一个数是j，表示i与j要好，第二个数是从i直接传递谣言到j所用的时间，数据的输出是两个整数，第一个为选点的散布谣言的起点，第二个整数时所有人得知谣言的最短时间

例如，对于数据1，

可知如果从3开始传播，则1，2得知谣言的时间都是2，所用的时间比从1，2开始传播所用的时间要短，故程序的输出时3 2；

参考代码：

 1 #include<iostream>
 2 #include<cstdlib>
 3 #include<cstdio>
 4 #include<cstring>
 5 #include<algorithm>
 6 #include<cmath>
 7 using namespace std;
 8 const int inf = (1 << 20);    //最大值
 9 int d[105][105];
10 struct prog
11 {
12     int x,y,d;    //定义起点，终点，所用时间
13 };
14 int main()
15 {
16     int n;
17     while(cin>>n,n)
18     {
19         int i,j;
20 
21         //先将路径初始化，让d[i][i]=0,d[i][j]=inf;i =/= j;
22         for(i=0;i<105;i++)
23         {
24             for(j=0;j<105;j++)
25             {
26                 if(i==j)
27                     d[i][j]=0;
28                 else
29                     d[i][j]=inf;
30             }
31         }
32 
33 
34         //根据输入的数据构造路径，注意题目的起点和终点范围是1……N，而数组是从0开始的，故要把所给的节点-1
35         for(i =0;i<n;i++)
36         {
37             int m;    //从i出发亲密人的个数
38             cin>>m;
39             for ( j = 0 ; j < m ; j ++)
40             {
41                 int a, b ;        //与i亲密的人及传递谣言的时间
42                 cin>>a>>b;
43                 d[i][a-1]=b;
44             }
45         }
46 
47     
48         //Floyd算法，你懂的
49         for ( int k = 0 ; k < n ; k ++ )
50         {
51             for(i = 0 ; i < n ; i ++)
52             {
53                 for ( j = 0 ; j < n ; j ++ )
54                 {
55                     if(d[i][j]>d[i][k]+d[k][j])
56                         d[i][j]=d[i][k]+d[k][j];
57                 }
58             }
59         }
60         
61         prog dij ,pij;//保存谣言散布的地点，终点，及时间，其中dij表示的是题目要求的，而pij表示的从固定点出发散布到某点的最长时间
62         dij.d=inf;//初始时将时间初始化为最大（因为要求的是从任意点出发让所有人知道谣言的最短时间）
63         for ( i = 0 ; i < n ; i ++ )
64         {
65             pij.d=0;//将从i散布谣言所用的时间初始化为0（因为此处要求的是从i出发散布谣言的最大时间）
66             for ( j = 0 ; j < n ; j ++ )
67             {
68                 if(pij.d<d[i][j])
69                 {//如果找到从i到j所用的时间要长与目前找到的从i出发用时的时间，则更新pij
70                     pij.d=d[i][j];
71                     pij.x=i;
72                     pij.y=j;
73                 }
74 
75             }
76             //如果从i出发用的时间比目前找到的用的最长时间还要短，则更新dij
77             if(dij.d>pij.d)
78                 dij=pij;
79         }
80         if(dij.d>=inf)//如果dij还一直是最大值，说明并不能让全部人得知谣言，则输出disjoint。但数据比较水，这个判断可省略 Orz
81             cout<<"disjoint"<<endl;
82         else//输出起点和时间，注意要将起点标号+1
83             cout<<dij.x+1<<' '<<dij.d<<endl;
84     }
85     return 0;
86 }