Currency Exchange--POJ 1860

1、题目类型：图论、最短路径、Bellman-Ford算法。

2、解题思路：Bellman-Ford算法简单应用。

3、注意事项：更新条件：d[j]<(d[i]-com[i][j])*rate[i][j]。

4、实现方法:

  
    

   
     
   
     #include
   
     <
   
     iostream
   
     >
   
     

   
     using
   
      
   
     namespace
   
      std;

   
     #define
   
      Max 110
   
     

   
     #define
   
      INF 99999999.9
   
     


   
     int
   
      n,m,s,flag;

   
     double
   
      v;

   
     double
   
      rate[Max][Max],com[Max][Max];


   
     void
   
      Bellman_ford()
{
 
   
     int
   
      i,j,k;
 
   
     double
   
      d[Max];
 
   
     for
   
      (i
   
     =
   
     0
   
     ;i
   
     <
   
     n;i
   
     ++
   
     ) 
 d[i]
   
     =-
   
     INF;
 s
   
     --
   
     ;
 d[s]
   
     =
   
     v;
 
   
     for
   
      (k
   
     =
   
     1
   
     ;k
   
     <
   
     n;k
   
     ++
   
     )
 
   
     for
   
      (i
   
     =
   
     0
   
     ;i
   
     <
   
     n;i
   
     ++
   
     )
 
   
     for
   
      (j
   
     =
   
     0
   
     ;j
   
     <
   
     n;j
   
     ++
   
     )
 
   
     if
   
      (d[j]
   
     <
   
     (d[i]
   
     -
   
     com[i][j])
   
     *
   
     rate[i][j]) 
 d[j]
   
     =
   
     (d[i]
   
     -
   
     com[i][j])
   
     *
   
     rate[i][j];

 flag
   
     =
   
     0
   
     ;
 
   
     for
   
      (i
   
     =
   
     0
   
     ;i
   
     <
   
     n;i
   
     ++
   
     )
 
   
     for
   
      (j
   
     =
   
     0
   
     ;j
   
     <
   
     n;j
   
     ++
   
     )
 
   
     if
   
      (d[j]
   
     <
   
     (d[i]
   
     -
   
     com[i][j])
   
     *
   
     rate[i][j]) 
 {
 flag
   
     =
   
     1
   
     ;
 
   
     return
   
     ;
 }
 
 
   
     if
   
     (d[s]
   
     >
   
     v)
 flag
   
     =
   
     1
   
     ;
}


   
     int
   
      main()
{
 
   
     int
   
      i,tmp1,tmp2;
 cin
   
     >>
   
     n
   
     >>
   
     m
   
     >>
   
     s
   
     >>
   
     v;
 
   
     for
   
     (i
   
     =
   
     0
   
     ;i
   
     <
   
     m;i
   
     ++
   
     )
 {
 cin
   
     >>
   
     tmp1
   
     >>
   
     tmp2;
 tmp1
   
     --
   
     ,tmp2
   
     --
   
     ;
 cin
   
     >>
   
     rate[tmp1][tmp2]
   
     >>
   
     com[tmp1][tmp2];
 cin
   
     >>
   
     rate[tmp2][tmp1]
   
     >>
   
     com[tmp2][tmp1];
 }
 Bellman_ford();
 
   
     if
   
     (flag)
 cout
   
     <<
   
     "
   
     YES
   
     "
   
     <<
   
     endl;
 
   
     else
   
     
 cout
   
     <<
   
     "
   
     NO
   
     "
   
     <<
   
     endl;
 
   
     return
   
      
   
     1
   
     ;
}