Wormholes--POJ 3259

1、题目类型：图论、最短路径、Bellman-Ford算法。

2、解题思路：（1）根据输入记录所有的边，其中无向边转换为有向边记录；（2）Bellman-Ford算法松弛所有的边，寻找是否存在负圈。

3、注意事项：注意Mov[]数组的大小。

4、实现方法:

  
    

   
     
   
     #include
   
     <
   
     iostream
   
     >
   
     

   
     using
   
      
   
     namespace
   
      std;

   
     #define
   
      Maxn 510
   
     

   
     #define
   
      Maxw 6000
   
     

   
     #define
   
      INF 99999
   
     


   
     struct
   
      TMove{
 
   
     int
   
      s,e,t;
};
TMove Mov[Maxw];

   
     int
   
      n,m,w,flag,cnt;


   
     void
   
      Bellman_ford()
{
 
   
     int
   
      i,j,start,end,time;
 
   
     double
   
      d[Maxn];
 
   
     for
   
      (i
   
     =
   
     0
   
     ;i
   
     <
   
     n;i
   
     ++
   
     ) 
 d[i]
   
     =
   
     INF;
 d[
   
     0
   
     ]
   
     =
   
     0
   
     ;
 
   
     for
   
      (i
   
     =
   
     1
   
     ;i
   
     <
   
     n;i
   
     ++
   
     )
 {
 
   
     for
   
      (j
   
     =
   
     0
   
     ;j
   
     <
   
     cnt;j
   
     ++
   
     )
 {
 start
   
     =
   
     Mov[j].s;
 end
   
     =
   
     Mov[j].e;
 time
   
     =
   
     Mov[j].t;
 
   
     if
   
      (d[end]
   
     >
   
     d[start]
   
     +
   
     time) 
 d[end]
   
     =
   
     d[start]
   
     +
   
     time;
 }
 }
 
   
     for
   
      (j
   
     =
   
     0
   
     ;j
   
     <
   
     cnt;j
   
     ++
   
     )
 {
 start
   
     =
   
     Mov[j].s;
 end
   
     =
   
     Mov[j].e;
 time
   
     =
   
     Mov[j].t;
 
   
     if
   
      (d[end]
   
     >
   
     d[start]
   
     +
   
     time) 
 {
 flag
   
     =
   
     1
   
     ;
 
   
     return
   
     ;
 }
 }
}


   
     int
   
      main()
{
 
   
     int
   
      i,f,s,e,t;
 cin
   
     >>
   
     f;
 
   
     while
   
     (f
   
     --
   
     )
 {
 cin
   
     >>
   
     n
   
     >>
   
     m
   
     >>
   
     w;
 memset(Mov,
   
     0
   
     ,
   
     sizeof
   
     (Mov));
 flag
   
     =
   
     0
   
     ,cnt
   
     =
   
     0
   
     ;
 
   
     for
   
     (i
   
     =
   
     0
   
     ;i
   
     <
   
     m;i
   
     ++
   
     )
 {
 cin
   
     >>
   
     s
   
     >>
   
     e
   
     >>
   
     t;
 s
   
     --
   
     ,e
   
     --
   
     ;
 Mov[cnt].s
   
     =
   
     s;
 Mov[cnt].e
   
     =
   
     e;
 Mov[cnt].t
   
     =
   
     t;
 cnt
   
     ++
   
     ;
 Mov[cnt].s
   
     =
   
     e;
 Mov[cnt].e
   
     =
   
     s;
 Mov[cnt].t
   
     =
   
     t;
 cnt
   
     ++
   
     ;
 }
 
   
     for
   
     (i
   
     =
   
     0
   
     ;i
   
     <
   
     w;i
   
     ++
   
     )
 {
 cin
   
     >>
   
     s
   
     >>
   
     e
   
     >>
   
     t;
 s
   
     --
   
     ,e
   
     --
   
     ;
 Mov[cnt].s
   
     =
   
     s;
 Mov[cnt].e
   
     =
   
     e;
 Mov[cnt].t
   
     =-
   
     t;
 cnt
   
     ++
   
     ; 
 } 
 Bellman_ford();
 
   
     if
   
     (flag)
 cout
   
     <<
   
     "
   
     YES
   
     "
   
     <<
   
     endl;
 
   
     else
   
     
 cout
   
     <<
   
     "
   
     NO
   
     "
   
     <<
   
     endl;
 }
 
   
     return
   
      
   
     1
   
     ;
}