POJ 1125 Stockbroker Grapevine

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.

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【题目分析】
是要求一个点，走到其他所有点的路径长度的和的最小值。由于有很多个源点，只能用floyd跑一边，然后再暴力的统计一下就好了。

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【代码】

#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>using namespace std;int n,m,b,c,dis[101][101];int main(){    while (cin>>n&&n)    {        memset(dis,0x3f,sizeof dis);        for (int i=1;i<=n;++i) dis[i][i]=0;        for (int i=1;i<=n;++i)        {            cin>>m;            for (int j=1;j<=m;++j)                cin>>b>>c,dis[i][b]=min(dis[i][b],c);        }        for (int k=1;k<=n;++k)            for (int i=1;i<=n;++i)                for (int j=1;j<=n;++j)                    dis[i][j]=min(dis[i][j],dis[i][k]+dis[k][j]);        int ans=0x3f3f3f3f,ai=-1;        for (int i=1;i<=n;++i)        {            int tmp=0;            for (int j=1;j<=n;++j)                tmp=max(dis[i][j],tmp);            if (tmp<ans) ai=i;            ans=min(ans,tmp);        }        if (ans==0x3f3f3f3f) cout<<"disjoint"<<endl;        else cout<<ai<<" "<<ans<<endl;    }}