Frogger--POJ 2253

1、题目类型：图论、最小生成树、Prim算法。

2、解题思路：Prim算法生成最小生成树过程中寻找最长Edge。

3、注意事项：每个case结束后，输出空行；pow()函数中需强制转换double。

4、实现方法:

  
    

   
     
   
     #include
   
     <
   
     iostream
   
     >
   
     
#include
   
     <
   
     cmath
   
     >
   
     
#include
   
     <
   
     algorithm
   
     >
   
     

   
     using
   
      
   
     namespace
   
      std;

   
     #define
   
      Max 210
   
     

   
     #define
   
      INF 9999999.9
   
     


   
     struct
   
      TNode{
 
   
     int
   
      x,y;
};
TNode Node[Max];

   
     int
   
      n,casenum,cnt;

   
     double
   
      map[Max][Max];

   
     bool
   
      vis[Max];


   
     double
   
      Prim()
{
 
   
     int
   
      i,j,pos;
 
   
     double
   
      dis[Max],ans
   
     =
   
     0
   
     ;
 memset(vis,
   
     0
   
     ,
   
     sizeof
   
     (vis));
 
   
     for
   
     (i
   
     =
   
     0
   
     ;i
   
     <
   
     n;i
   
     ++
   
     )
 dis[i]
   
     =
   
     map[
   
     0
   
     ][i];
 vis[
   
     0
   
     ]
   
     =
   
     1
   
     ;
 
   
     for
   
     (i
   
     =
   
     0
   
     ;i
   
     <
   
     n
   
     -
   
     1
   
     ;i
   
     ++
   
     )
 {
 pos
   
     =
   
     min_element(dis
   
     +
   
     1
   
     ,dis
   
     +
   
     n)
   
     -
   
     dis;
 
   
     if
   
     (ans
   
     <
   
     dis[pos])
 ans
   
     =
   
     dis[pos];
 
   
     if
   
     (pos
   
     ==
   
     1
   
     )
 
   
     break
   
     ;
 
   
     else
   
     
 {
 vis[pos]
   
     =
   
     1
   
     ;
 dis[pos]
   
     =
   
     INF;
 
   
     for
   
     (j
   
     =
   
     0
   
     ;j
   
     <
   
     n;j
   
     ++
   
     )
 {
 
   
     if
   
     (
   
     !
   
     vis[j]
   
     &&
   
     dis[j]
   
     >
   
     map[pos][j])
 dis[j]
   
     =
   
     map[pos][j];
 }
 }
 }
 
   
     return
   
      ans;
}


   
     int
   
      main()
{
 
   
     int
   
      i,j;
 
   
     while
   
     (cin
   
     >>
   
     n 
   
     &&
   
      n)
 {
 casenum
   
     ++
   
     ;
 
   
     for
   
     (i
   
     =
   
     0
   
     ;i
   
     <
   
     n;i
   
     ++
   
     )
 {
 cin
   
     >>
   
     Node[i].x
   
     >>
   
     Node[i].y;
 }
 
   
     for
   
     (i
   
     =
   
     0
   
     ;i
   
     <
   
     n;i
   
     ++
   
     )
 {
 map[i][i]
   
     =
   
     INF;
 
   
     for
   
     (j
   
     =
   
     i
   
     +
   
     1
   
     ;j
   
     <
   
     n;j
   
     ++
   
     )
 {
 map[i][j]
   
     =
   
     map[j][i]
   
     =
   
     sqrt(pow((
   
     double
   
     )(Node[i].x
   
     -
   
     Node[j].x),
   
     2
   
     )
   
     +
   
     pow((
   
     double
   
     )(Node[i].y
   
     -
   
     Node[j].y),
   
     2
   
     ));
 }
 }
 cout
   
     <<
   
     "
   
     Scenario #
   
     "
   
     <<
   
     casenum
   
     <<
   
     endl;
 printf(
   
     "
   
     Frog Distance = %.3f\n
   
     "
   
     ,Prim());
 cout
   
     <<
   
     endl;
 }
 
   
     return
   
      
   
     1
   
     ;
}