poj 1125 floyd算法求最短路

题目：

Stockbroker Grapevine

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 34641 Accepted: 19201

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

32 2 4 3 52 1 2 3 62 1 2 2 253 4 4 2 8 5 31 5 84 1 6 4 10 2 7 5 202 2 5 1 50

Sample Output

3 23 10

N表示共N个点，然后底下N行，第i行第一个数字t表示后面有t个pair，每个pair两个数字分别代表x, w，i点到x点距离w，注意这是单向道路。题目求从哪个点当作起点可以使得该点走到距离该点最远的距离为最短。假设i点作起点，走到距离i点最远的那个点距离为Di，求所有Di中最小的那个值，输出i及Di。另外题目要求这个i点需要能够到达所有其他的点。

分析：

先用floyd求出任意两点之间的最短路，然后对于每个顶点求出它到达其他点的最大距离（每一行的最大值），求所有最大距离中的最小值。如果最小值是inf说明从任何一个顶点出发都无法到达其他所有点，输出disjoint。

题目特点：题意不好理解，数据太弱。可以用来熟悉floyd。

代码：

#include <stdio.h>#include <iostream>using namespace std;const int inf=99999999;int gra[105][105];int n,t;inline void init(){for(int i=1;i<=n;++i){for(int j=1;j<=n;++j){gra[i][j]=(i==j)?0:inf;//初始化要留心 }}}inline void floyd(){for(int k=1;k<=n;++k){for(int i=1;i<=n;++i){for(int j=1;j<=n;++j){gra[i][j]=min(gra[i][j],gra[i][k]+gra[k][j]);}}}}int main(){int a,b;while(~scanf("%d",&n),n){init();for(int i=1;i<=n;++i){scanf("%d",&t);for(int j=0;j<t;++j){scanf("%d%d",&a,&b);gra[i][a]=b;}}floyd(); int mind=inf,maxd,tt;for(int i=1;i<=n;++i){maxd=0;for(int j=1;j<=n;++j){maxd=max(maxd,gra[i][j]);}if(maxd<mind){mind=maxd;tt=i;}}if(mind==inf) printf("disjoint\n");else printf("%d %d\n",tt,mind);}return 0;}