| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 28231 | Accepted: 15659 |
Description
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
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
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
#include<stdio.h>
#include<string.h>
const int inf = 0x3f3f3f3f , M = ;
int d[M] ;
int map[M][M] ;
int n ; int main ()
{
//freopen ("a.txt" , "r" , stdin) ;
int x , v , t ;
while (~ scanf ("%d" , &n)) {
if (n == )
break ;
for (int i = ; i <= n ; i++) {
for (int j = ; j <= n ; j++) {
map[i][j] = inf ;
}
}
for (int i = ; i <= n ; i++) {
scanf ("%d" , &x) ;
while (x--) {
scanf ("%d%d" , &v , &t) ;
map[i][v] = t ;
}
}
for (int k = ; k <= n ; k++) {
for (int i = ; i <= n ; i++) {
for (int j = ; j<= n ; j++) {
if (map[i][k] + map[k][j] < map[i][j] && i != j)
map[i][j] = map[i][k] + map[k][j] ;
}
}
}
int maxn ;
for (int i = ; i <= n ; i++) {
maxn = - ;
for (int j = ; j <= n ; j++) {
if (i != j && maxn < map[i][j])
maxn = map[i][j] ;
}
d[i] = maxn ;
}
int minn = inf, k;
for (int i = ; i <= n ; i++) {
if (minn > d[i]) {
minn = d[i] ;
k = i ;
}
}
if (minn == inf)
puts ("disjoint") ;
else
printf ("%d %d\n" , k , minn) ;
}
return ;
}
floyd算法写法上很简单，适合多源最短路径发