Stockbroker Grapevine--POJ 1125

1、题目类型：图论、最短路径、Floyd算法。

2、解题思路：Floyd算法的简单应用

3、注意事项：注意n为0的特殊情况。

4、实现方法:

  
    

   
     
   
     #include
   
     <
   
     iostream
   
     >
   
     

   
     #define
   
      inf 9999999
   
     
#include
   
     <
   
     algorithm
   
     >
   
     

   
     using
   
      
   
     namespace
   
      std;


   
     int
   
      n,map[
   
     110
   
     ][
   
     110
   
     ];


   
     void
   
      Floyd()
{
 
   
     int
   
      i,j,k,dis[
   
     110
   
     ][
   
     110
   
     ];
 memset(dis,
   
     0
   
     ,
   
     sizeof
   
     (dis));
 
   
     for
   
     (i
   
     =
   
     1
   
     ;i
   
     <=
   
     n;i
   
     ++
   
     )
 {
 
   
     for
   
     (j
   
     =
   
     1
   
     ;j
   
     <=
   
     n;j
   
     ++
   
     )
 dis[i][j]
   
     =
   
     inf;
 }
 
   
     for
   
     (i
   
     =
   
     1
   
     ;i
   
     <=
   
     n;i
   
     ++
   
     )
 {
 
   
     for
   
     (j
   
     =
   
     1
   
     ;j
   
     <=
   
     n;j
   
     ++
   
     )
 {
 
   
     if
   
     (i
   
     !=
   
     j)
 dis[i][j]
   
     =
   
     map[i][j];
 }
 }
 
   
     for
   
     (k
   
     =
   
     1
   
     ;k
   
     <=
   
     n;k
   
     ++
   
     )
 {
 
   
     for
   
     (i
   
     =
   
     1
   
     ;i
   
     <=
   
     n;i
   
     ++
   
     )
 
   
     for
   
     (j
   
     =
   
     1
   
     ;j
   
     <=
   
     n;j
   
     ++
   
     )
 {
 
   
     if
   
     (dis[i][j]
   
     >
   
     dis[i][k]
   
     +
   
     dis[k][j])
 dis[i][j]
   
     =
   
     dis[i][k]
   
     +
   
     dis[k][j];
 }
 }
 
   
     int
   
      ans
   
     =-
   
     1
   
     ,pos
   
     =
   
     0
   
     ,arr[
   
     110
   
     ];
 
   
     for
   
     (i
   
     =
   
     1
   
     ;i
   
     <=
   
     n;i
   
     ++
   
     )
 {
 ans
   
     =-
   
     1
   
     ;
 
   
     for
   
     (j
   
     =
   
     1
   
     ;j
   
     <=
   
     n;j
   
     ++
   
     )
 {
 
   
     if
   
     (i
   
     !=
   
     j)
 
   
     if
   
     (ans
   
     <
   
     dis[i][j])
 {
 ans
   
     =
   
     dis[i][j];
 pos
   
     =
   
     i;
 } 
 }
 arr[pos]
   
     =
   
     ans;
 }
 pos
   
     =
   
     min_element(arr
   
     +
   
     1
   
     ,arr
   
     +
   
     n
   
     +
   
     1
   
     )
   
     -
   
     arr;
 
   
     if
   
     (
   
     *
   
     min_element(arr
   
     +
   
     1
   
     ,arr
   
     +
   
     n
   
     +
   
     1
   
     )
   
     ==
   
     inf)
 cout
   
     <<
   
     "
   
     disjoint
   
     "
   
     <<
   
     endl;
 
   
     else
   
     
 cout
   
     <<
   
     pos
   
     <<
   
     '
   
      
   
     '
   
     <<*
   
     min_element(arr
   
     +
   
     1
   
     ,arr
   
     +
   
     n
   
     +
   
     1
   
     )
   
     <<
   
     endl;
}

   
     int
   
      main()
{
 
   
     while
   
     (cin
   
     >>
   
     n 
   
     &&
   
      n)
 {
 
   
     int
   
      i,j,m,a,t;
 
   
     for
   
     (i
   
     =
   
     1
   
     ;i
   
     <=
   
     n;i
   
     ++
   
     )
 
   
     for
   
     (j
   
     =
   
     1
   
     ;j
   
     <=
   
     n;j
   
     ++
   
     )
 map[i][j]
   
     =
   
     inf;
 
   
     for
   
     (i
   
     =
   
     1
   
     ;i
   
     <=
   
     n;i
   
     ++
   
     )
 {
 cin
   
     >>
   
     m;
 
   
     for
   
     (j
   
     =
   
     0
   
     ;j
   
     <
   
     m;j
   
     ++
   
     )
 {
 cin
   
     >>
   
     a
   
     >>
   
     t;
 map[i][a]
   
     =
   
     t;
 }
 }
 
   
     if
   
     (n
   
     ==
   
     1
   
     )
 {
 cout
   
     <<
   
     n
   
     <<
   
     '
   
      
   
     '
   
     <<
   
     '
   
     0
   
     '
   
     <<
   
     endl;
 
   
     continue
   
     ;
 }
 Floyd();
 }
 
   
     return
   
      
   
     0
   
     ;
}