哪一点的位置权值最小——sgu114. Telecasting station

列方程可知，最小点一定在给定的某点上

先排序

后枚举计算某点的权值

枚举可以由前面的推出后面的，不必重复计算

View Code

   
     

    
      #include
    
      <
    
      stdio.h
    
      >
    
      
#include
    
      <
    
      iostream
    
      >
    
      
#include
    
      <
    
      algorithm
    
      >
    
      

    
      using
    
       
    
      namespace
    
       std;


    
      struct
    
       data
{
 
    
      double
    
       x;
 
    
      int
    
       p;
}map[
    
      15099
    
      ];


    
      int
    
       cmp(data a,data b)
{
 
    
      return
    
       a.x
    
      <
    
      b.x;
}


    
      int
    
       main()
{
 
    
      int
    
       n;
 scanf(
    
      "
    
      %d
    
      "
    
      ,
    
      &
    
      n);

 
    
      int
    
       i;
 
    
      for
    
      (i
    
      =
    
      0
    
      ;i
    
      <
    
      n;i
    
      ++
    
      )
 {
 scanf(
    
      "
    
      %lf %d
    
      "
    
      ,
    
      &
    
      map[i].x,
    
      &
    
      map[i].p);
 }

 
    
      if
    
      (n
    
      ==
    
      1
    
      )
 {
 printf(
    
      "
    
      %.5lf\n
    
      "
    
      ,map[
    
      0
    
      ].x);
 
    
      return
    
       
    
      0
    
      ;
 }
 sort(
    
      &
    
      map[
    
      0
    
      ],
    
      &
    
      map[n],cmp);

 
    
      double
    
       min
    
      =
    
      0
    
      ;
 
    
      double
    
       rx
    
      =
    
      map[
    
      0
    
      ].x,lp
    
      =
    
      0
    
      ,rp
    
      =
    
      map[
    
      0
    
      ].p;
 
    
      for
    
      (i
    
      =
    
      1
    
      ;i
    
      <
    
      n;i
    
      ++
    
      )
 {
 min
    
      +=
    
      (map[i].x
    
      -
    
      rx)
    
      *
    
      map[i].p;
 rp
    
      +=
    
      map[i].p;
 }

 
    
      double
    
       add
    
      =
    
      min,cha;
 
    
      for
    
      (i
    
      =
    
      1
    
      ;i
    
      <
    
      n;i
    
      ++
    
      )
 {
 lp
    
      +=
    
      map[i
    
      -
    
      1
    
      ].p;
 rp
    
      -=
    
      map[i
    
      -
    
      1
    
      ].p;

 cha
    
      =
    
      map[i].x
    
      -
    
      map[i
    
      -
    
      1
    
      ].x;
 add
    
      =
    
      add
    
      -
    
      cha
    
      *
    
      rp
    
      +
    
      cha
    
      *
    
      lp;
 
    
      if
    
      (add
    
      <
    
      min)
 {
 rx
    
      =
    
      map[i].x;
 min
    
      =
    
      add;
 }
 }

 printf(
    
      "
    
      %.5lf\n
    
      "
    
      ,rx);
}