Stockbroker Grapevine
| Time Limit: 1000MS |
|
Memory Limit: 10000K |
| Total Submissions: 18729 |
|
Accepted: 10120 |
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
构图，求出从一个顶点到其余顶点最远的距离，然后求每个顶点这样做之后的最小值，即为最小时间
显然求多源最短路，题目中给出的数据范围是1-100，floyd正合适，O（N^3）
写好之后交了两边之后都wa，不明白
后来自习看了半天，发现找完最短路f[][]时候，查找答案过程中是在map[][]中找的，坏习惯害死人啊，
这里说一下，其实只用一个数组记录图就行，在起基础上操作即可
这次的样例够阴险的，找完最短路后矩阵居然没变……
1
#include<stdio.h>
2#include<string.h>
3#include<math.h>
4#define MAX 100
5#define M 210000000
6int point,ans;
7int n,map[MAX+5][MAX+5],f[MAX+5][MAX+5];
8void init()
9
{
10
int i,j,t;
11 int a,b;
12 for (i=0; i<=n ; i++ )
13 for (j=0; j<=n ; j++ )
14
{
15 map[i][j]=M;
16
}
17 for (i=1; i<=n ; i++ )
18 {
19 scanf("%d",&t);
20 for (j=1; j<=t; j++)
21 {
22 scanf("%d%d",&a,&b);
23 map[i][a]=b;
24 }
25 }
26
27 for (i=1; i<=n ; i++ )
28 for (j=1; j<=n ; j++ )
29 f[i][j]=map[i][j];
30
}
31int work()
32{
33 int min;
34 int i,j,k;
35 for (k=1; k<=n ; k++ )
36 for (i=1; i<=n ; i++ )
37 for (j=1; j<=n ; j++ )
38 if ((i!=j)&&(j!=k)&&(k!=i))
39 if (f[i][j]>f[i][k]+f[k][j])
40 {
41 f[i][j]=f[i][k]+f[k][j];
42 }
43 ans=M;
44 for (i=1; i<=n ; i++ )
45 {
46 min=0;
47 for (j=1; j<=n ; j++ )
48 if (f[i][j]>min&&i!=j)
49 {
50 min=f[i][j];
51 }
52 if (min<M)
53 {
54 if (min<ans)
55 {
56 point=i;
57 ans=min;
58 }
59 }
60 }
61}
62int main()
63{
64 int flag;
65 int i;
66 while (scanf("%d",&n)!=EOF&&n!=0)
67 {
68 init();
69 work();
70 if (ans>M)
71 printf("disjoint\n");
72 else
73 printf("%d %d\n",point,ans);
74 }
75 return 0;
76}
77