POJ 2253 Frogger

解题思路:最短路径dijkstra算法

  
    

   
     
   
     #include
   
     <
   
     iostream
   
     >
   
     
#include 
   
     <
   
     cmath
   
     >
   
     
#include 
   
     <
   
     iomanip
   
     >
   
     

   
     using
   
      
   
     namespace
   
      std;


   
     #define
   
      MaxNum 1e8 
   
     

   
     int
   
      main()
{
 
   
     int
   
      n, start, tempS;
 
   
     bool
   
      visited[
   
     205
   
     ];
 
   
     double
   
      dist[
   
     205
   
     ];
 
   
     double
   
      minDist, tempDist;
 
   
     int
   
      cases 
   
     =
   
      
   
     0
   
     ;
 
   
     struct
   
      
 {
 
   
     int
   
      x, y;
 }p[
   
     205
   
     ];

 
   
     while
   
      (cin 
   
     >>
   
      n 
   
     &&
   
      n)
 {
 cases
   
     ++
   
     ;
 memset(visited, 
   
     0
   
     , 
   
     sizeof
   
     (
   
     bool
   
     ) 
   
     *
   
      
   
     205
   
     );
 fill(
   
     &
   
     dist[
   
     0
   
     ], 
   
     &
   
     dist[
   
     204
   
     ], MaxNum);

 
   
     for
   
      (
   
     int
   
      i 
   
     =
   
      
   
     0
   
     ; i 
   
     <
   
      n; i
   
     ++
   
     )
 cin 
   
     >>
   
      p[i].x 
   
     >>
   
      p[i].y;

 visited[
   
     0
   
     ] 
   
     =
   
      
   
     true
   
     ;
 dist[
   
     0
   
     ] 
   
     =
   
      
   
     0
   
     ;
 start 
   
     =
   
      
   
     0
   
     ;

 
   
     for
   
      (
   
     int
   
      i 
   
     =
   
      
   
     0
   
     ; i 
   
     <
   
      n; i
   
     ++
   
     )
 {
 minDist 
   
     =
   
      MaxNum;
 
   
     for
   
      (
   
     int
   
      j 
   
     =
   
      
   
     0
   
     ; j 
   
     <
   
      n; j
   
     ++
   
     )
 {
 
   
     if
   
      (
   
     !
   
     visited[j])
 {
 tempDist 
   
     =
   
      pow(
   
     double
   
     (p[start].x 
   
     -
   
      p[j].x), 
   
     2
   
     );
 tempDist 
   
     +=
   
      pow(
   
     double
   
     (p[start].y 
   
     -
   
      p[j].y), 
   
     2
   
     );
 tempDist 
   
     =
   
      sqrt(tempDist);
 tempDist 
   
     =
   
      max(tempDist, dist[start]);
 dist[j] 
   
     =
   
      min(dist[j], tempDist);

 
   
     if
   
      (minDist 
   
     >
   
      dist[j])
 {
 minDist 
   
     =
   
      dist[j];
 tempS 
   
     =
   
      j;
 }
 }
 }
 start 
   
     =
   
      tempS;
 visited[start] 
   
     =
   
      
   
     true
   
     ;
 
   
     if
   
      (start 
   
     ==
   
      
   
     1
   
     )
 
   
     break
   
     ;
 }
 
   
     if
   
      (cases 
   
     !=
   
      
   
     1
   
     )
 cout 
   
     <<
   
      
   
     "
   
     \n
   
     "
   
     ;
 cout 
   
     <<
   
      
   
     "
   
     Scenario #
   
     "
   
      
   
     <<
   
      cases 
   
     <<
   
      endl;
 cout 
   
     <<
   
      
   
     "
   
     Frog Distance = 
   
     "
   
      
   
     <<
   
      
   
     fixed
   
      
   
     <<
   
      setprecision(
   
     3
   
     ) 
   
     <<
   
      dist[
   
     1
   
     ] 
   
     <<
   
      endl;
 }
 
   
     return
   
      
   
     0
   
     ;
}