POJ 1125 Stockbroker Grapevine

解题思路: 首先Floyd算法求出所有顶点之间的最小距离，计算最小最大距离

  
    

   
     
   
     #include 
   
     <
   
     iostream
   
     >
   
     

   
     using
   
      
   
     namespace
   
      std;


   
     #define
   
      MaxNum 1e8
   
     


   
     int
   
      main()
{
 
   
     int
   
      p[
   
     105
   
     ][
   
     105
   
     ];
 
   
     int
   
      dist[
   
     105
   
     ];
 
   
     int
   
      n, m, v, t, start, minTime;
 
   
     while
   
      (cin 
   
     >>
   
      n 
   
     &&
   
      n)
 {
 
 fill(
   
     &
   
     p[
   
     0
   
     ][
   
     0
   
     ], 
   
     &
   
     p[
   
     104
   
     ][
   
     104
   
     ], MaxNum);
 
   
     for
   
      (
   
     int
   
      i 
   
     =
   
      
   
     1
   
     ; i 
   
     <=
   
      n; i
   
     ++
   
     )
 {
 p[i][i] 
   
     =
   
      
   
     0
   
     ;
 cin 
   
     >>
   
      m;
 
   
     while
   
     (m
   
     --
   
     )
 {
 cin 
   
     >>
   
      v 
   
     >>
   
      t;
 p[i][v] 
   
     =
   
      t;
 }
 }

 
   
     for
   
      (
   
     int
   
      k 
   
     =
   
      
   
     1
   
     ; k 
   
     <=
   
      n; k
   
     ++
   
     )
 
   
     for
   
      (
   
     int
   
      i 
   
     =
   
      
   
     0
   
     ; i 
   
     <=
   
      n; i
   
     ++
   
     )
 
   
     for
   
      (
   
     int
   
      j 
   
     =
   
      
   
     0
   
     ; j 
   
     <=
   
      n; j
   
     ++
   
     )
 
   
     if
   
      (p[i][j] 
   
     >
   
      p[i][k] 
   
     +
   
      p[k][j])
 p[i][j] 
   
     =
   
      p[i][k] 
   
     +
   
      p[k][j];

 minTime 
   
     =
   
      MaxNum;
 
   
     for
   
      (
   
     int
   
      i 
   
     =
   
      
   
     1
   
     ; i 
   
     <=
   
      n; i
   
     ++
   
     )
 {
 
   
     int
   
      maxTime 
   
     =
   
      
   
     0
   
     ;
 
   
     for
   
      (
   
     int
   
      j 
   
     =
   
      
   
     1
   
     ; j 
   
     <=
   
      n; j
   
     ++
   
     )
 
   
     if
   
      (maxTime 
   
     <
   
      p[i][j])
 maxTime 
   
     =
   
      p[i][j];
 
   
     if
   
     (minTime 
   
     >
   
      maxTime)
 {
 minTime 
   
     =
   
      maxTime;
 start 
   
     =
   
      i;
 }
 }
 
   
     if
   
      (minTime 
   
     ==
   
      MaxNum)
 cout 
   
     <<
   
      
   
     "
   
     disjoint
   
     "
   
     <<
   
     endl;
 
   
     else
   
     
 cout 
   
     <<
   
      start
   
     <<
   
      
   
     "
   
      
   
     "
   
      
   
     <<
   
      minTime 
   
     <<
   
      endl;
 }
 
   
     return
   
      
   
     0
   
     ;
}