PKU1125 最短路径 Floyd-Warshall算法

Stockbroker Grapevine

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 14288		Accepted: 7689

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

 

 

简单图论题
,Floyd
求最短路径之后统计
.

代码如下
:

#include<stdio.h>
#include<string.h>
#define MAXN 105
#define inf 100000000
int main ()
{
int n,i,j,k,nn,v,min,max,mini;
int d[MAXN][MAXN];
while (scanf("%d",&n)&&n)
{
memset(d,0,sizeof(d));
for (i=1; i<=n; ++i)
{
scanf("%d",&nn);
for (j=1; j<=nn; ++j)
{
scanf("%d%d",&k,&v);
d[i][k]=v;
}
}
for (i=1; i<=n; ++i)
{
for (j=1; j<=n; ++j)
{
if (d[i][j]==0) d[i][j]=inf;
}
}
for (k=1; k<=n; ++k)
{
for (i=1; i<=n; ++i)
{
for (j=1; j<=n; ++j)
{
if (i!=j)
{
int temp=d[i][k]+d[k][j];
if (temp<d[i][j]) d[i][j]=temp;
}
}
}
}
min=2100000000;
for (i=1; i<=n; ++i)
{
max=0;
for (j=1; j<=n; ++j)
{
if (d[i][j]==inf&&i!=j) {max=0;break;}
if (d[i][j]>max&&i!=j) max=d[i][j];
}
if (max<min&&max!=0)
{
min=max;
mini=i;
}
}
if (min!=2100000000) printf("%d %d/n",mini,min);
else printf("disjoint/n");
}
return 0;
}