poj1125 Stockbroker Grapevine

Stockbroker Grapevine

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 33800 Accepted: 18666

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

这个题就是说，给每个人编号，告诉你从一个人到另一个人消息传播的时间，然后让你求一个最快的传播中所花费的最少时间。

这个地方要主要的就是，消息传播给每个人是同时的，也就是1传播给2的时间是3分钟，而1传播给3的时间是5分钟，那么5分钟之后，2,3都能收到这个消息了。

我们可以把每个人之间传播的时间大小当做权值，题目中给出的路径很明显是单向的。

总的思路就是从1-n每次当做原点搜一个最短路，其中到每个点距离中最大的便是这次传播的值，然后从这些点中找一个最小的值便是答案。

这种每个点都搜一次的不用想，floyd肯定比spfa快..

#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>#include <string>#include <cmath>using namespace std;const int inf=0x3f3f3f;int tu[110][110];int n;void floyd(){    int i,j,k;    for(k=1; k<=n; ++k)        for(i=1; i<=n; ++i)            for(j=1; j<=n; ++j)            {                if(tu[i][j]>tu[i][k]+tu[k][j])                {                    tu[i][j]=tu[i][k]+tu[k][j];                }            }    int mm,pos;    int ans=inf;    for(i=1; i<=n; ++i)    {        mm=-1;        for(j=1; j<=n; ++j)        {            mm=max(mm,tu[i][j]);        }        if(ans>mm)        {            ans=mm;            pos=i;        }    }    if(ans>=inf)printf("disjoint\n");    else printf("%d %d\n",pos,ans);}int main(){    int y,w;    int i,j;    while(~scanf("%d",&n))    {        if(!n)break;        int k;        for(i=1; i<=n; ++i)            for(j=1; j<=n; ++j)            {                if(i==j)tu[i][j]=0;                else tu[i][j]=inf;            }        for(i=1; i<=n; ++i)        {            scanf("%d",&k);            while(k--)            {                scanf("%d%d",&y,&w);                tu[i][y]=w;            }        }        floyd();    }    return 0;}