题意:对于一个点i,设f(i)=max{mindis[i,j]} (j≠i).

其中mindis是各个点对之间的最短路.求min{f(i)} (1<=i<=n).

分析:floyd求出最短路即可.

code:

var   person,time,n,i,j,k,p,min,max,mini,minperson:longint;      map:array[0..110,0..110] of longint;      f:boolean;begin      readln(n);      while n<>0 do      begin            fillchar(map,sizeof(map),1);            for i:=1 to n do            begin                  read(p);                  for j:=1 to p do                  begin                        read(person,time);                        map[i,person]:=time;                  end;                  readln;            end;            for k:=1 to n do               for i:=1 to n do                  for j:=1 to n do                  if map[i,k]+map[k,j]
     
      max then max:=map[i,j];                  end;                  if f then                    if max