pku 3259 Wormholes

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 8968		Accepted: 3171

      本题的意思是说在农场里有wormhole可以是时光倒流，现在给你一个地图，问是否能够运用这些wormhole从而使得当FJ回到原处发点的时候比第一次出发的时候要早。

代码：

 

  
    

   
     
   
      1 
   
     #include
   
     <
   
     stdio.h
   
     >
   
     

   
      2 
   
      
   
     struct
   
      node

   
      3 
   
     {

   
      4 
   
      
   
     int
   
      a,b,t;

   
      5 
   
     }s[
   
     6000
   
     ];

   
      6 
   
      
   
     int
   
      dist[
   
     505
   
     ],n,m,w;

   
      7 
   
     
   
     int
   
      relax(
   
     int
   
      u,
   
     int
   
      v,
   
     int
   
      t)

   
      8 
   
     {

   
      9 
   
      
   
     if
   
     (dist[v]
   
     >
   
     dist[u]
   
     +
   
     t)

   
     10 
   
      {

   
     11 
   
      dist[v]
   
     =
   
     dist[u]
   
     +
   
     t;

   
     12 
   
      
   
     return
   
      
   
     1
   
     ;

   
     13 
   
      }

   
     14 
   
      
   
     return
   
      
   
     0
   
     ;

   
     15 
   
     }

   
     16 
   
     
   
     int
   
      bellman(
   
     int
   
      v0)

   
     17 
   
     {

   
     18 
   
      
   
     int
   
      i,j;

   
     19 
   
      
   
     for
   
     (i
   
     =
   
     1
   
     ;i
   
     <=
   
     n;i
   
     ++
   
     )

   
     20 
   
      dist[i]
   
     =
   
     0xfffffff
   
     ;

   
     21 
   
      dist[v0]
   
     =
   
     0
   
     ;

   
     22 
   
      
   
     int
   
      flag;

   
     23 
   
      
   
     for
   
     (i
   
     =
   
     1
   
     ;i
   
     <
   
     n;i
   
     ++
   
     )

   
     24 
   
      {

   
     25 
   
      flag
   
     =
   
     0
   
     ;

   
     26 
   
      
   
     for
   
     (j
   
     =
   
     1
   
     ;j
   
     <=
   
     2
   
     *
   
     m
   
     +
   
     w;j
   
     ++
   
     )

   
     27 
   
      {

   
     28 
   
      
   
     if
   
     (relax(s[j].a,s[j].b,s[j].t))

   
     29 
   
      {

   
     30 
   
      flag
   
     =
   
     1
   
     ;

   
     31 
   
      }

   
     32 
   
      }

   
     33 
   
      
   
     if
   
     (
   
     !
   
     flag)

   
     34 
   
      
   
     break
   
     ;

   
     35 
   
      }

   
     36 
   
      
   
     for
   
     (j
   
     =
   
     1
   
     ;j
   
     <=
   
     2
   
     *
   
     m
   
     +
   
     w;j
   
     ++
   
     )

   
     37 
   
      
   
     if
   
     (relax(s[j].a,s[j].b,s[j].t))

   
     38 
   
      
   
     return
   
      
   
     1
   
     ;

   
     39 
   
      
   
     return
   
      
   
     0
   
     ;

   
     40 
   
     }

   
     41 
   
     
   
     int
   
      main()

   
     42 
   
     {

   
     43 
   
      
   
     int
   
      f,i,a,b,t;

   
     44 
   
      scanf(
   
     "
   
     %d
   
     "
   
     ,
   
     &
   
     f);

   
     45 
   
      
   
     while
   
     (f
   
     --
   
     )

   
     46 
   
      {

   
     47 
   
      scanf(
   
     "
   
     %d%d%d
   
     "
   
     ,
   
     &
   
     n,
   
     &
   
     m,
   
     &
   
     w);

   
     48 
   
      
   
     for
   
     (i
   
     =
   
     1
   
     ;i
   
     <=
   
     m;i
   
     ++
   
     )

   
     49 
   
      {

   
     50 
   
      scanf(
   
     "
   
     %d%d%d
   
     "
   
     ,
   
     &
   
     a,
   
     &
   
     b,
   
     &
   
     t);

   
     51 
   
      s[i].a
   
     =
   
     a;

   
     52 
   
      s[i].b
   
     =
   
     b;

   
     53 
   
      s[i].t
   
     =
   
     t;

   
     54 
   
      s[i
   
     +
   
     m].a
   
     =
   
     b;

   
     55 
   
      s[i
   
     +
   
     m].b
   
     =
   
     a;

   
     56 
   
      s[i
   
     +
   
     m].t
   
     =
   
     t;

   
     57 
   
      }

   
     58 
   
      
   
     for
   
     (i
   
     =
   
     1
   
     ;i
   
     <=
   
     w;i
   
     ++
   
     )

   
     59 
   
      {

   
     60 
   
      scanf(
   
     "
   
     %d%d%d
   
     "
   
     ,
   
     &
   
     a,
   
     &
   
     b,
   
     &
   
     t);

   
     61 
   
      s[m
   
     *
   
     2
   
     +
   
     i].a
   
     =
   
     a;

   
     62 
   
      s[m
   
     *
   
     2
   
     +
   
     i].b
   
     =
   
     b;

   
     63 
   
      s[m
   
     *
   
     2
   
     +
   
     i].t
   
     =-
   
     t;

   
     64 
   
      }

   
     65 
   
      
   
     if
   
     (bellman(
   
     1
   
     ))

   
     66 
   
      printf(
   
     "
   
     YES\n
   
     "
   
     );

   
     67 
   
      
   
     else
   
     

   
     68 
   
      printf(
   
     "
   
     NO\n
   
     "
   
     );

   
     69 
   
      }

   
     70 
   
      
   
     return
   
      
   
     0
   
     ;

   
     71 
   
     }

   
     72 
   
     

   
     73