POJ 2253 Prim算法及测试数据

prim算法求最小生成树中的最大边

测试数据：

  
    

   
     2
   
     

   
     0
   
      
   
     0
   
     

   
     3
   
      
   
     4
   
     


   
     3
   
     

   
     17
   
      
   
     4
   
     

   
     19
   
      
   
     4
   
     

   
     18
   
      
   
     5
   
     

   
     8
   
     

   
     1
   
      
   
     1
   
     

   
     4
   
      
   
     0
   
     

   
     1
   
      
   
     2
   
     

   
     2
   
      
   
     2
   
     

   
     3
   
      
   
     2
   
     

   
     4
   
      
   
     2
   
     

   
     3
   
      
   
     0
   
     

   
     5
   
      
   
     1
   
     

   
     3
   
     

   
     9
   
      
   
     10
   
     

   
     10
   
      
   
     10
   
     

   
     100
   
      
   
     10
   
     

   
     6
   
     

   
     5
   
      
   
     5
   
     

   
     100
   
      
   
     100
   
     

   
     4
   
      
   
     4
   
     

   
     3
   
      
   
     3
   
     

   
     2
   
      
   
     2
   
     

   
     1
   
      
   
     1
   
     

   
     5
   
     

   
     1
   
      
   
     2
   
     

   
     2
   
      
   
     1
   
     

   
     3
   
      
   
     2
   
     

   
     4
   
      
   
     1
   
     

   
     5
   
      
   
     2
   
     

   
     3
   
     

   
     999
   
      
   
     999
   
     

   
     1
   
      
   
     1
   
     

   
     3
   
      
   
     3
   
     

   
     0
  
    

结果

  
    

   
     Scenario #
   
     1
   
     
Frog Distance 
   
     =
   
      
   
     5.000
   
     

Scenario #
   
     2
   
     
Frog Distance 
   
     =
   
      
   
     1.414
   
     

Scenario #
   
     3
   
     
Frog Distance 
   
     =
   
      
   
     1.414
   
     

Scenario #
   
     4
   
     
Frog Distance 
   
     =
   
      
   
     1.000
   
     

Scenario #
   
     5
   
     
Frog Distance 
   
     =
   
      
   
     134.350
   
     

Scenario #
   
     6
   
     
Frog Distance 
   
     =
   
      
   
     1.414
   
     

Scenario #
   
     7
   
     
Frog Distance 
   
     =
   
      
   
     1408.557
  
    

  
    

   
     #include 
   
     <
   
     iostream
   
     >
   
     
#include 
   
     <
   
     math.h
   
     >
   
     

   
     #define
   
      MAX 201
   
     


   
     #define
   
      sq(a) ((a)*(a)) 
   
     

   
     float
   
      dis[MAX][MAX];
typedef 
   
     struct
   
      node{
 
   
     int
   
      x,y;
 
}Node;

   
     long
   
      
   
     int
   
      INF 
   
     =
   
     99999999
   
     ;

   
     bool
   
      inTree[MAX];
Node nodes[MAX];

   
     void
   
      ptable(
   
     int
   
      count){
 
   
     int
   
      i,j;
 
   
     for
   
     (i
   
     =
   
     0
   
     ;i
   
     <
   
     count;i
   
     ++
   
     ){
 
   
     for
   
     (j
   
     =
   
     0
   
     ;j
   
     <
   
     count;j
   
     ++
   
     )
 printf(
   
     "
   
     (%d,%d)%10f
   
     "
   
     ,i,j,dis[i][j]);
 printf(
   
     "
   
     \n
   
     "
   
     );
 }
}

   
     bool
   
      allInTree(
   
     int
   
      count){

 
   
     for
   
     (
   
     int
   
      i
   
     =
   
     0
   
     ;i
   
     <
   
     count;i
   
     ++
   
     )
 
   
     if
   
     (
   
     !
   
     inTree[i])
   
     return
   
      
   
     false
   
     ;
 
   
     return
   
      
   
     true
   
     ;
}

   
     int
   
      prim(
   
     int
   
      v,
   
     int
   
      n){
 
   
     int
   
      i,j,ti,tj;
 
   
     float
   
      mindis,maxdis 
   
     =
   
      
   
     -
   
     1
   
     ;
 memset(inTree,
   
     false
   
     ,
   
     sizeof
   
     (inTree));
 inTree[
   
     0
   
     ] 
   
     =
   
      
   
     true
   
     ;
 
   
     while
   
     (
   
     !
   
     allInTree(n)){
 mindis
   
     =
   
     INF;

 
   
     //
   
     找到最小权值
   
     

   
      
   
     for
   
     (i
   
     =
   
     0
   
     ;i
   
     <
   
     n;i
   
     ++
   
     )
 {
 
   
     if
   
     (inTree[i]){
 
   
     for
   
     (j
   
     =
   
     0
   
     ;j
   
     <
   
     n;j
   
     ++
   
     )
 
   
     if
   
     (i
   
     !=
   
     j
   
     &&!
   
     inTree[j]
   
     &&
   
     dis[i][j]
   
     <=
   
     mindis){
 mindis
   
     =
   
     dis[i][j];
 ti 
   
     =
   
      i;
 tj 
   
     =
   
      j;
 }
 }
 }
 
   
     if
   
     (mindis
   
     >
   
     maxdis)maxdis
   
     =
   
     mindis;
 
   
     //
   
     找到终点
   
     

   
      
   
     if
   
     (ti
   
     ==
   
     1
   
     ||
   
     tj
   
     ==
   
     1
   
     ){
 
 
   
     return
   
      maxdis;
 }
 
   
     //
   
     printf("%d %d\nIntree:",ti,tj);
 
   
     //
   
     inTree[ti]=true;
   
     

   
      inTree[tj]
   
     =
   
     true
   
     ;

 
   
     //
   
     for(int k=0;k<n;k++)
 
   
     //
   
     if(inTree[k])printf("%d ",k);
 
   
     //
   
     printf("\n");
 
   
     //
   
     ptable(n);
   
     

   
      }
 
 
 

 
   
     return
   
      maxdis;
}

   
     int
   
      main(
   
     int
   
      argc, 
   
     char
   
     *
   
      argv[])
{
 
   
     //
   
     freopen("i:
   
     //
   
     t.txt","r",stdin);
 
   
     //
   
     freopen("i:
   
     //
   
     out.txt","w",stdout);
   
     

   
      
   
     int
   
      count,i,j,c 
   
     =
   
      
   
     1
   
     ;
 
   
     while
   
     (scanf(
   
     "
   
     %d
   
     "
   
     ,
   
     &
   
     count)
   
     &&
   
     count){
 
 
 
 
   
     for
   
     (i
   
     =
   
     0
   
     ;i
   
     <
   
     count;i
   
     ++
   
     )
 {
 scanf(
   
     "
   
     %d%d
   
     "
   
     ,
   
     &
   
     nodes[i].x,
   
     &
   
     nodes[i].y);
 
   
     for
   
     (j
   
     =
   
     0
   
     ;j
   
     <
   
     i;j
   
     ++
   
     ){
 dis[i][j]
   
     =
   
     sq(nodes[i].x
   
     -
   
     nodes[j].x)
   
     +
   
     sq(nodes[i].y
   
     -
   
     nodes[j].y);
 dis[j][i]
   
     =
   
     dis[i][j];
 }
 dis[i][i] 
   
     =
   
      
   
     0
   
     ;
 } 
 printf(
   
     "
   
     Scenario #%d\n
   
     "
   
     ,c
   
     ++
   
     );
 printf(
   
     "
   
     Frog Distance = %.3lf\n\n
   
     "
   
     ,sqrt((
   
     float
   
     )prim(
   
     0
   
     ,count)));
 


 
 }
 
   
     return
   
      
   
     0
   
     ;
}