TOJ 2857 Stockbroker Grapevine

描述

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.

输入

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.

输出

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.

样例输入

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

样例输出

3 2
3 10

题目来源

Southern African 2001

求一个Stockbroker把消息传递给其他Stockbroker所用的最短时间。

floyd求出两点之间消息传递的时间后，再求最短时间。

#include <stdio.h>
#define MAXN 105
#define inf 0x3f3f3f3f

int N;
int map[MAXN][MAXN];
int ans[MAXN];

void floyd(){
    for(int k=; k<=N; k++){
        for(int i=; i<=N; i++){
            for(int j=; j<=N; j++){
                if(i==k||i==j)continue;
                if(map[i][k]!=inf && map[k][j]!=inf &&
                        map[i][k]+map[k][j]<map[i][j]){
                    map[i][j]=map[i][k]+map[k][j];
                }
            }
        }
    }
}
int main(int argc, char *argv[])
{
    int t,u,v,w;
    while(scanf("%d",&N)!=EOF && N){
        for(int i=; i<=N; i++){
            for(int j=; j<=N; j++){
                if(i==j)map[i][j]=;
                else map[i][j]=inf;
            }
        }
        for(int u=; u<=N; u++){
            scanf("%d",&t);
            while(t--){
                scanf("%d %d",&v,&w);
                map[u][v]=w;
            }
            ans[u]=-;
        }
        floyd();
        for(int i=; i<=N; i++){
            for(int j=; j<=N; j++){
                if(map[i][j]>ans[i])ans[i]=map[i][j];
            }
        }
        int min=inf,index;
        for(int i=; i<=N; i++){
            if(ans[i]<min){
                min=ans[i];
                index=i;
            }
        }
        if(min==inf){
            printf("disjoint\n");
        }else{
            printf("%d %d\n",index,min);
        }
    }
    return ;
}