Acm hdu 1162 Eddy's picture (simple Minimum Spanning Tree, start to use the template)

Eddy's picture

Time Limit: 2000/1000 MS (Java/others) memory limit: 65536/32768 K (Java/Others)
Total submission (s): 2454 accepted submission (s): 1184

Problem descriptioneddy begins to like painting pictures recently, he is sure of himself to become a painter. every day Eddy draws pictures in his small room, and he usually puts out his newest pictures to let his friends appreciate. but the result it can be imagined, the friends are not interested in his picture. eddy feels very puzzled, in order to change all friends's view to his technical of painting pictures, So Eddy creates a problem for the his friends of you.
Problem descriptions as follows: given you some coordinates pionts on a drawing paper, every point links with the ink with the straight line, causes all points finally to link in the same place. how many distants does your duty discover the shortest length which the ink draws?

Inputthe first line contains 0 <n <= 100, the number of point. for each point, a line follows; each following line contains two real numbers indicating the (x, y) coordinates of the point.

Input contains multiple test cases. process to the end of file.

Outputyour program prints a single real number to two decimal places: the minimum total length of ink lines that can connect all the points.

Sample input3
1.0 1.0
2.0 2.0
2.0 4.0

Sample output3.41

Authoreddy

Recommendjgshining

 /*  
Hdu1162 Eddy's picture
Given n vertices, find the shortest line segment to connect all vertices. A simple Minimum Spanning Tree
  */  
# Include  <  Stdio. h  >  
# Include  <  Math. h  >  
# Include <  String  . H  >  
  # Define  Maxn 110  
  Struct  Node  //  Definition point  
  {
  Double  X, Y;
} Node [maxn];
  Double Cost [maxn] [maxn];  //  Consumption Matrix  
  Double  Distance (node A, Node B)  //  Distance between two points  
  {
  Return  SQRT (A. x  -  B. X)  *  (A. x  - B. X)  +  (A. Y  -  B. Y)  *  (A. Y  -  B. Y ));
}
  //  **************************************** *******************************
  //  Apply a prim Template
  //  **************************************** ******************************* 
  # Define  Typec double  
  Const  Typec INF  =  0x3f3f3f  ;
  Int  Vis [maxn];
Typec lowc [maxn];
Typec prim (typec cost [] [maxn],  Int  N)  //  Point from 0 ~ N-1  
 {
  Int  I, j, P;
Typec minc, res  =  0  ;
Memset (VIS,  0  ,  Sizeof  (VIS ));
Vis [  0  ]  =  1  ;
  For  (I =  1  ; I  <  N; I  ++  ) Lowc [I]  =  Cost [  0  ] [I];
  For  (I  =  1  ; I  <  N; I ++  )
{
Minc  =  INF;
P  =-  1  ;
  For  (J  =  0  ; J  <  N; j  ++  )
  If (Vis [J]  =  0  &&  Minc  >  Lowc [J])
{Minc  =  Lowc [J]; P  =  J ;}
  If  (Minc  =  INF)  Return    -  1  ;  //  Source image not connected  
  Res  + =  Minc; vis [p]  =  1  ;
  For  (J  =  0  ; J  < N; j  ++  )
  If  (Vis [J]  =  0  &&  Lowc [J]  >  Cost [p] [J])
Lowc [J]  =  Cost [p] [J];
}
  Return  Res;
}
  // **************************************** ****************************  
  Int  Main ()
{
  Int  I, j, N;
  While  (Scanf (  "  % D  "  ,  &  N)  ! =  EOF)
{
 For  (I  =  0  ; I  <  N; I  ++  )
Scanf (  "  % Lf  "  ,  &  Node [I]. X,  &  Node [I]. y );
 For  (I  =  0  ; I  <  N; I  ++  )
  For  (J  =  0  ; J  <  N; j  ++  )
{
 If  (I  =  J) cost [I] [J]  =  0  ;
  Else  
Cost [I] [J]  =  Distance (node [I], node [J]);
}
Printf (  "  %. 2lf \ n  "  , Prim (cost, n ));
}
  Return    0  ;
}