(最短路-floyd+邻接矩阵)PKU-1125 Stockbroker Grapevine

原题链接 http://poj.org/problem?id=1125
Stockbroker Grapevine

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

问题：为达到某些目的，商家会散布一些假信息。但是每个人只相信他们认为可靠的人那里获取信息。一个消息从消息源转到下面有一定的时间。一个消息的持续时间就是它从开始传播到无法传播为止的最长时间。现给出一个消息网络以及每条渠道所用时间，在网络中选择一个人作为消息源，使得消息的持续时间最短。

思路：用Floyd算法求每点之间的最短距离，一个点到其他点的最大距离就是该点的传播时间。找出最小即可。

Source Code

/*
最短路径算法：floyd_warshall
*/
#include <iostream>
using namespace std;

#define Max 105
#define INF 10000000

int map[Max][Max];
int result[Max][Max];
int pre[Max][Max];

//多源最短路径,floyd_warshall算法,复杂度O(n^3)
//求出所有点对之间的最短路经,传入图的大小points 和邻接阵map[][]
//返回各点间最短距离result[]和路径pre[],pre[i][j]记录i到j最短路径上j的父结点
//可更改路权类型,路权必须非负!
void floyd_warshall(int points){
int i,j,k;
for (i = 0; i < points; i++) {
for (j = 0; j < points; j++) {
result[i][j] = map[i][j];
pre[i][j] = (i == j) ? -1 : i;
}
}

//探测点一定要放在最前面(WA多次)
for (k = 0; k < points; k++) {
for (i = 0; i < points; i++) {
for (j = 0; j < points; j++) {
if (i != j && (result[i][k] + result[k][j] < result[i][j])) {
result[i][j] = result[i][k] + result[k][j];
pre[i][j]=pre[k][j];
}
}
}
}
}

int main() {
int points, lines, end, len;
int i, j;
while (scanf("%d", &points) !=EOF) {

if (points == 0){
break;
}

for (i = 0; i <= points; i++) {
for (j = 0; j <= points; j++) {
map[i][j] = INF;
}
}

for (i = 1; i <= points; i++) {
scanf("%d", &lines);
for (j = 0; j < lines; j++) {
scanf("%d %d", &end, &len);
map[i - 1][end - 1] = len;
}
}

floyd_warshall(points);

int index = -1;
int MaxTime, MinTime;
MinTime = INF;
for (i = 0; i < points; i++) {
MaxTime = -1;
for (j = 0; j < points; j++) {
if (i == j) {
continue;
}
if (result[i][j] > MaxTime) {
MaxTime = result[i][j];
}
}
if (MinTime > MaxTime) {
MinTime = MaxTime;
index = i;
}
}

if (MinTime == INF) {
printf("disjoint\n");
} else {
printf("%d %d\n", index + 1, MinTime);
}
}

}