poj 1125 Stockbroker Grapevine （Floyd最短路）【模板】

tockbrokers 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

 【题解】 

 大致题意：

 老板要给全体员工发通知，想尽快让每位员工都收到通知，现在有n位员工，每个人都和其他一些人有联系，（也存在有人和任何人都没联系的情况），一个人可以和与他有联系的人互发短信，a和b互发短信的时间可能是不一样的，现在想知道从哪个员工开始发短信，可以使所有的人都收到通知，并输出所有人都收到短信的最小总时间。如果有人通知不到，输出 disjoint。

 分析：

 弗洛伊德算法，专门解决这类最短路问题，网上有这个算法的详细原理，百度一下，你就知道~

 

 【AC代码】

 

#include<cstdio>#include<iostream>#include<algorithm>#include<cstring>using namespace std;const int N=105;const int inf=0x3f3f3f3f;int mapp[N][N];int main(){    int m,n;    int a,b;    while(~scanf("%d",&m),m)    {        for(int i=1;i<=m;++i)        {            mapp[i][i]=0;            for(int j=1;j<i;++j)                mapp[i][j]=mapp[j][i]=inf;//初始化所有点间距为无穷大        }        for(int i=1;i<=m;++i)        {            scanf("%d",&n);            while(n--)            {                scanf("%d%d",&a,&b);                mapp[i][a]=b;            }        }        for(int i=1;i<=m;++i)//弗洛伊德最短路松弛            for(int j=1;j<=m;++j)                for(int k=1;k<=m;++k)                if(mapp[j][k]>mapp[j][i]+mapp[i][k])                mapp[j][k]=mapp[j][i]+mapp[i][k];        int minn=inf;        int p=0;        for(int i=1;i<=m;++i)        {            int maxn=0;            for(int j=1;j<=m;++j)//找到从第i个人到最后一个人的最大时间花费            {                if(i==j) continue;                if(maxn<mapp[i][j])                    maxn=mapp[i][j];            }            if(maxn<minn)//更新花费时间的最大值            {                minn=maxn;                p=i;            }        }        if(minn==inf)//如果最大值不从在  说明有未连通的点            printf("disjoint\n");        else            printf("%d %d\n",p,minn);    }    return 0;}