Stockbroker Grapevine

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 20000/10000K (Java/Other)

Total Submission(s) : 20   Accepted Submission(s) : 6

Problem 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

 

#include <map>#include <set>#include <list>#include <queue>#include <stack>#include <cmath>#include <ctime>#include <vector>#include <bitset>#include <cstdio>#include <string>#include <numeric>#include <cstring>#include <cstdlib>#include <iostream>#include <algorithm>#include <functional>using namespace std;typedef long long  ll;typedef unsigned long long ull;int dx[4]= {-1,1,0,0};int dy[4]= {0,0,-1,1}; //up down left rightbool inmap(int x,int y,int n,int m){    if(x<1||x>n||y<1||y>m)return false;    return true;}int hashmap(int x,int y,int m){    return (x-1)*m+y;}#define eps 1e-8#define inf 0x7fffffff#define debug puts("BUG")#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1#define read freopen("in.txt","r",stdin)#define write freopen("out.txt","w",stdout)#define maxn 200int s[maxn][maxn];int n;//情报工作= =!/***    问从哪个人开始传递谣言,会使所有人都知道谣言的时间最短,    所有人都知道谣言的时间,就是从某人向其他人传播时,最晚获得谣言的那个人的时间~~    注意,会有人根本收不到谣言~~        求所有人的最短路~~        收不到谣言的特判~~***/void Floyd(){    //cout<<dist[0][0]<<endl;    for(int i=1; i<=n; i++)        for(int j=1; j<=n; j++)            for(int k=1; k<=n; k++)            {                if(j==k)                {                    s[j][k]=0;                    continue;                }                if(s[j][k]>s[j][i]+s[i][k])                    s[j][k]=s[j][i]+s[i][k];            }}void print(){    int mini=0,mins=16843009;    for(int j=1; j<=n; j++)    {        int maxs=-1,flag=1;        for(int k=1; k<=n; k++)        {            if(s[j][k]==s[0][0])            {flag=0;continue;}            if(s[j][k]>maxs)                maxs=s[j][k];        }        if(maxs<mins&&flag)        {            mini=j;            mins=maxs;        }    }    if(mins==16843009)//收不到谣言,也就是说,最短时间一直是正无穷~~    {        printf("disjoint\n");        return ;    }    printf("%d %d\n",mini,mins);}int main(){    while(scanf("%d",&n)!=EOF&&n)    {        memset(s,1,sizeof(s));        for(int i=1; i<=n; i++)        {            int st=i,en,M;            scanf("%d",&M);            for(int m=0; m<M; m++)            {                scanf("%d",&en);                scanf("%d",&s[st][en]);            }            s[st][st]=0;        }        Floyd();        print();    }    return 0;}