poj 2253 Frogger

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 9968		Accepted: 3327

     本题就是求最短路中最长的那一段，prim算法当U=2是即可跳出

代码：

 

  
    

   
     
   
      1 
   
     #include
   
     <
   
     stdio.h
   
     >
   
     

   
      2 
   
     #include
   
     <
   
     math.h
   
     >
   
     

   
      3 
   
     

   
      4 
   
      
   
     int
   
      x[
   
     205
   
     ],y[
   
     205
   
     ],max,n;

   
      5 
   
      
   
     void
   
      prim(
   
     int
   
      a,
   
     int
   
      b)

   
      6 
   
     {

   
      7 
   
      
   
     int
   
      visit[
   
     205
   
     ],dist[
   
     205
   
     ],min,u,i,j,distance;

   
      8 
   
      
   
     for
   
     (i
   
     =
   
     1
   
     ;i
   
     <=
   
     n;i
   
     ++
   
     )

   
      9 
   
      {

   
     10 
   
      dist[i]
   
     =
   
     (x[i]
   
     -
   
     a)
   
     *
   
     (x[i]
   
     -
   
     a)
   
     +
   
     (y[i]
   
     -
   
     b)
   
     *
   
     (y[i]
   
     -
   
     b);

   
     11 
   
      visit[i]
   
     =
   
     0
   
     ;

   
     12 
   
      }

   
     13 
   
      visit[
   
     1
   
     ]
   
     =
   
     1
   
     ;

   
     14 
   
      
   
     for
   
     (i
   
     =
   
     1
   
     ;i
   
     <
   
     n;i
   
     ++
   
     )

   
     15 
   
      {

   
     16 
   
      min
   
     =
   
     0xfffffff
   
     ;

   
     17 
   
      
   
     for
   
     (j
   
     =
   
     1
   
     ;j
   
     <=
   
     n;j
   
     ++
   
     )

   
     18 
   
      {

   
     19 
   
      
   
     if
   
     (
   
     !
   
     visit[j] 
   
     &&
   
      dist[j]
   
     <
   
     min)

   
     20 
   
      {

   
     21 
   
      u
   
     =
   
     j;

   
     22 
   
      min
   
     =
   
     dist[j];

   
     23 
   
      }

   
     24 
   
      }

   
     25 
   
      
   
     if
   
     (min
   
     >
   
     max)

   
     26 
   
      max
   
     =
   
     min;

   
     27 
   
      
   
     if
   
     (u
   
     ==
   
     2
   
     )

   
     28 
   
      
   
     return
   
      ;

   
     29 
   
      visit[u]
   
     =
   
     1
   
     ;

   
     30 
   
      
   
     for
   
     (j
   
     =
   
     1
   
     ;j
   
     <=
   
     n;j
   
     ++
   
     )

   
     31 
   
      {

   
     32 
   
      distance
   
     =
   
     (x[u]
   
     -
   
     x[j])
   
     *
   
     (x[u]
   
     -
   
     x[j])
   
     +
   
     (y[u]
   
     -
   
     y[j])
   
     *
   
     (y[u]
   
     -
   
     y[j]);

   
     33 
   
      
   
     if
   
     (
   
     !
   
     visit[j] 
   
     &&
   
      distance
   
     <
   
     dist[j])

   
     34 
   
      {

   
     35 
   
      dist[j]
   
     =
   
     distance;

   
     36 
   
      }

   
     37 
   
      }

   
     38 
   
      }

   
     39 
   
     }

   
     40 
   
      
   
     int
   
      main()

   
     41 
   
     {

   
     42 
   
      
   
     int
   
      count
   
     =
   
     0
   
     ,i;

   
     43 
   
      
   
     while
   
     (scanf(
   
     "
   
     %d
   
     "
   
     ,
   
     &
   
     n)
   
     !=
   
     EOF)

   
     44 
   
      {

   
     45 
   
      max
   
     =
   
     0
   
     ;

   
     46 
   
      count
   
     ++
   
     ;

   
     47 
   
      
   
     if
   
     (n
   
     ==
   
     0
   
     )

   
     48 
   
      
   
     break
   
     ;

   
     49 
   
      
   
     for
   
     (i
   
     =
   
     1
   
     ;i
   
     <=
   
     n;i
   
     ++
   
     )

   
     50 
   
      {

   
     51 
   
      scanf(
   
     "
   
     %d%d
   
     "
   
     ,
   
     &
   
     x[i],
   
     &
   
     y[i]);

   
     52 
   
      }

   
     53 
   
      prim(x[
   
     1
   
     ],y[
   
     1
   
     ]);

   
     54 
   
      printf(
   
     "
   
     Scenario #%d\n
   
     "
   
     ,count);

   
     55 
   
      printf(
   
     "
   
     Frog Distance = %.3f\n
   
     "
   
     ,sqrt(max
   
     *
   
     1.0
   
     ));

   
     56 
   
      printf(
   
     "
   
     \n
   
     "
   
     );

   
     57 
   
      }

   
     58 
   
      
   
     return
   
      
   
     0
   
     ;

   
     59 
   
     }

   
     60