POJ 1125 可不可能遍历所有点情况下的最短路径

Stockbroker Grapevine

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 33859		Accepted: 18711

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

题意：

首先，题目可能有多组测试数据，每个测试数据的第一行为经纪人数量N（当N=0时，输入数据结束），然后接下来N行描述第i（1<=i<=N）个经纪人与其他经纪人的关系（教你如何画图）。
每行开头数字M为该行对应的经纪人有多少个经纪人朋友（该节点的出度，可以为0），然后紧接着M对整数，每对整数表示成a,b,则表明该经纪人向第a个经纪人传递信息需要b单位时间
（即第i号结点到第a号结点的孤长为b），整张图为有向图，即弧Vij 可能不等于弧Vji（数据很明显，这里是废话）。当构图完毕后，求当从该图中某点出发，将“消息”传播到整个经
纪人网络的最小时间，输出这个经纪人号和最小时间。最小时间的判定方式为——从这个经纪人（结点）出发，整个经纪人网络中最后一个人接到消息的时。如果有一个或一个以上经纪
人无论如何无法收到消息，输出“disjoint”

#include 
#include 
#include 
using namespace std;
const int inf = 0x3f3f3f3f;
const int N = 110;
int map[N][N];
int dis[N][N];
int n;
void Floyd()
{
    //Floyd算法必须初始化！
    for(int i = 1;i<=n;i++)
        for(int j =1;j<=n;j++)
        if(i==j) map[i][j]=0;
    else map[i][j]=inf;

    for(int i=1; i<=n; i++)
    {
        int t,x,s;
        scanf("%d",&t);
        while(t--)
        {
            scanf("%d%d",&x,&s);
            map[i][x]=s;
        }
    }
    for(int t=1; t<=n; t++)
        for(int i=1; i<=n; i++)
            for(int j=1; j<=n; j++)
                map[i][j]=min(map[i][j],map[i][t]+map[t][j]);
}
int main()
{
    while(cin>>n && n)
    {
        Floyd();
        int ans=0x3f3f3f3f;
        int time;
        for(int i=1;i<=n;i++)
        {
            int tmp=0;
            for(int j=1;j<=n;j++)
            {   //在j里面找最小的
                if(map[i][j]>tmp)
                    tmp=map[i][j];
            }
            if(ans>tmp){
                //在最后的j中找出最小的i
                ans=tmp;
                time=i;
            }

        }
        if(ans>=inf)
            printf("%s","disjoint");
        else
        printf("%d %d\n",time,ans);
    }
    return 0;
}

转载于:https://www.cnblogs.com/mingrigongchang/p/6246251.html