Stockbroker Grapevine – GCC – O(n3)解法

Contents

1 原题
2 解法
- 2.1 代码
- 2.2 运算过程

原题

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

Source

Southern African 2001

解法

1. 用Floyd算法求出每个人到其他所有人的最短时间。
2. 找出最快的那个人。
(我不会描述，看最后的过程吧～)

代码

/******************************************
 * This is a solution related to poj.org/problem?id=1125
 * Input and Output are specified in the above page.
 *
 * Author: feichashao@gmail.com
 * Modify: 2014.12.7
 *
 *****************************************/

#include 

/* Define constants */
#define INF 2000  // Means unreachable.
#define MAX_BROKERS 100 // The maximum number of Stockbrokers. Used to constract distance array.

/* Declare global variables */
// Distance matrix of Floyd.
int distance[MAX_BROKERS+1][MAX_BROKERS+1] = { INF };
int num_brokers;
int best_person;

/* Declare functions */
void Init_distance();
void Read_distance();
void Floyd_distance();
void Find_person();
#ifdef DEBUG
void Print_distance();
#endif


int main()
{
    while(1){
    // Read number of brokers.
    // if equals 0, terminate.
    scanf("%d", &num_brokers);
    if( !num_brokers ) return 0;
    
    // Constract initial distance matrix.
    Init_distance();
    // Read distance from stdin.
    Read_distance();
    // Calculat using Floyd algorithm
    Floyd_distance();
    // Find out the fastest person and time.
    Find_person();
    // Print result.
    printf("%d %d\n", best_person, distance[best_person][0]);
    }
    return 0;
}

void Init_distance(){
    if(num_brokers < 1 || num_brokers > 100 ) {
        printf("Number of brokers invalid.");
        return ;
    }

    int i, j;
    for( i = 1; i <= num_brokers; i++ ) {
        for ( j = 1; j <= num_brokers; j++ ) {
            distance[i][j] = INF;
        }
        distance[i][i] = 0; // zero to itself.
    }

#ifdef DEBUG
    printf("\n\n\n-------Init Distance----------\n");
    Print_distance();
#endif
}

void Read_distance(){
    int i,j;
    int pairs;
    int contact, time;
    for( i = 1; i <= num_brokers; i++ ) {
        scanf("%d", &pairs);
        for( j = 0; j < pairs; j++){
            scanf("%d %d", &contact, &time);
            distance[i][contact] = time;
        }
    }

#ifdef DEBUG
    printf("\n\n\n-------Read Distance----------\n");
    Print_distance();
#endif
}

void Floyd_distance(){
    int i, j, k;
    for ( i = 1; i <= num_brokers; i++ ) {
        for ( j = 1; j <= num_brokers; j++ ) {
            for( k = 1; k <= num_brokers; k++ ) {
                // If transfer by i is cheaper, modify the distance matrix.
                if( distance[j][k] > distance[j][i] + distance[i][k] ) {
                    distance[j][k] = distance[j][i] + distance[i][k] ;
                }
            }
        }
#ifdef DEBUG
        printf("\n\n------- After #%d operation------\n", i);
        Print_distance();
#endif
    }
}

/******
 * Find the fastest person.
 * Put the longest time of each person cost to spread in the first col.
 * Like distance[i][0] records the longest time cost i to spread the disinformation.
 * ****/
void Find_person(){
    best_person = 0; // Init value.
    distance[0][0] = INF;
    int i,j;
    for( i = 1; i <= num_brokers; i++ ) {
        distance[i][0] = distance[i][1];
        for( j = 2; j <= num_brokers; j++ ) {
            if( distance[i][j] > distance[i][0] ) distance[i][0] = distance[i][j];
        }
        if( distance[i][0] <= distance[best_person][0] ) {
            best_person = i;
        }
#ifdef DEBUG
        printf("Longest distance of %d is %d\n", i, distance[i][0]);
#endif
    }
}


#ifdef DEBUG
void Print_distance(){
    int i,j;
    for( i = 1; i <= num_brokers; i++ ) {
        for( j = 1; j <= num_brokers; j++ ) {
            if(distance[i][j] >= INF){
                printf("I ");
            } else {
                printf("%d ", distance[i][j]);
            }
        }
        printf("\n\n");
    }
}
#endif

运算过程

feichashao@Aspire-4743:~/poj$ gcc -o Stockbroker_debug Stockborker.c -D DEBUG
feichashao@Aspire-4743:~/poj$ ./Stockbroker_debug 
3



-------Init Distance----------
0 I I 

I 0 I 

I I 0 

2 2 4 3 5
2 1 2 3 6
2 1 2 2 2



-------Read Distance----------
0 4 5 

2 0 6 

2 2 0 



------- After #1 operation------
0 4 5 

2 0 6 

2 2 0 



------- After #2 operation------
0 4 5 

2 0 6 

2 2 0 



------- After #3 operation------
0 4 5 

2 0 6 

2 2 0 

Longest distance of 1 is 5
Longest distance of 2 is 6
Longest distance of 3 is 2
3 2