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floyd多源最短路径

#include<iostream>#include<cstring>#include<cstdio>#define inf 20 using namespace std; int dis[110][110];int i,j,k,n;void floyd(){    for(k=1; k<=n; ++k)        for(i=1; i<=n; ++i)            for(j=1; j<=n; ++j)                if(i!=j && dis[i][j] > dis[i][k] +dis[k][j])                    dis[i][j]=dis[i][k]+dis[k][j];    int maxlen;    int min_max=ainf;    int flag;    for(i=1; i<=n; ++i)    {        maxlen=0;        for(j=1; j<=n; ++j)            if(i!=j && maxlen<dis[i][j])                maxlen=dis[i][j];        if(min_max>maxlen)        {            min_max=maxlen;            flag=i;        }    }    if(min_max<inf)        cout<<flag<<' '<<min_max<<endl;    else        cout<<"disjoint"<<endl;}int main(){    while(cin>>n&&n)    {        memset(dis,inf,sizeof(dis));        for(i=1; i<=n; ++i)        {            int m;            cin>>m;            for(j=1; j<=m; ++j)            {                int cat,time;                cin>>cat>>time;                dis[i][cat]=time;            }        }        floyd();    }}

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