这题的思路还是比较简单，用弗洛伊德算法打表后，枚举来找到最小值

代码如下 注意最后判断时候的语句 在这里错误了很多次

# include
     
      # include
      
       using namespace std;int p[105][105];const int INF = 99999999;int n;void floyd(){    for (int k = 1; k <= n; k++)        for (int i = 1; i <= n; i++)            for (int j = 1; j <= n; j++)                p[i][j] = min(p[i][j],p[i][k]+p[k][j]);}int main(){    int m, e, t;    while (cin >> n)    {        if (n == 0)            break;        for (int i = 1; i <= n; i++)            for (int k = 1; k <= n; k++)            {                if (i == k)                    p[i][k] = 0;                else                    p[i][k] = INF;            }        for (int i = 1; i <= n; i++)        {            cin >> m;            if (m)            {                for (int k = 1; k <= m; k++)                {                    cin >> e >> t;                    p[i][e] = t;                }            }        }        floyd();        int mymin = INF;        bool flag = false;        int s,j;        for (int i = 1; i <= n; i++)        {            int mymax = -INF;//注意这个初始化写在循环里            for ( j = 1; j <= n; j++)            {                if (i != j)                {                    if (p[i][j] == INF)                        break;                    mymax = max(mymax, p[i][j]);                }                            }            if (j == n + 1)                flag = true;            else                continue;                            if (mymin > mymax)                {                    mymin = mymax;                    s = i;                }        //上面这一部分的判断条件要想清楚        }        if (flag)            cout << s << " " << mymin << endl;        else            cout << "disjoint" << endl;    }    return 0;}