poj2785

大空间大时间的题，把4个数字分成两份，分别两两求和，得到两个长度n*n的一维数组，排序后比较进行匹配即可。

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      #include 
    
      <
    
      iostream
    
      >
    
      
#include 
    
      <
    
      cstdio
    
      >
    
      
#include 
    
      <
    
      cstdlib
    
      >
    
      
#include 
    
      <
    
      cstring
    
      >
    
      
#include 
    
      <
    
      algorithm
    
      >
    
      

    
      using
    
       
    
      namespace
    
       std;


    
      #define
    
       maxn 4004
    
      


    
      int
    
       n;

    
      int
    
       f1[maxn 
    
      *
    
       maxn];

    
      int
    
       f2[maxn 
    
      *
    
       maxn];

    
      int
    
       a[maxn], b[maxn], c[maxn], d[maxn];


    
      int
    
       main()
{
 
    
      //
    
      freopen("t.txt", "r", stdin);
    
      

    
       scanf(
    
      "
    
      %d
    
      "
    
      , 
    
      &
    
      n);
 
    
      for
    
       (
    
      int
    
       i 
    
      =
    
       
    
      0
    
      ; i 
    
      <
    
       n; i
    
      ++
    
      )
 scanf(
    
      "
    
      %d%d%d%d
    
      "
    
      , 
    
      &
    
      a[i], 
    
      &
    
      b[i], 
    
      &
    
      c[i], 
    
      &
    
      d[i]);
 
    
      for
    
       (
    
      int
    
       i 
    
      =
    
       
    
      0
    
      ; i 
    
      <
    
       n; i
    
      ++
    
      )
 
    
      for
    
       (
    
      int
    
       j 
    
      =
    
       
    
      0
    
      ; j 
    
      <
    
       n; j
    
      ++
    
      )
 f1[i 
    
      *
    
       n 
    
      +
    
       j] 
    
      =
    
       a[i] 
    
      +
    
       b[j];
 
    
      for
    
       (
    
      int
    
       i 
    
      =
    
       
    
      0
    
      ; i 
    
      <
    
       n; i
    
      ++
    
      )
 
    
      for
    
       (
    
      int
    
       j 
    
      =
    
       
    
      0
    
      ; j 
    
      <
    
       n; j
    
      ++
    
      )
 f2[i 
    
      *
    
       n 
    
      +
    
       j] 
    
      =
    
       c[i] 
    
      +
    
       d[j];
 sort(f1, f1 
    
      +
    
       n 
    
      *
    
       n);
 sort(f2, f2 
    
      +
    
       n 
    
      *
    
       n);
 
    
      int
    
       r 
    
      =
    
       n 
    
      *
    
       n 
    
      -
    
       
    
      1
    
      ;
 
    
      int
    
       ans 
    
      =
    
       
    
      0
    
      ;
 
    
      for
    
       (
    
      int
    
       i 
    
      =
    
       
    
      0
    
      ; i 
    
      <
    
       n 
    
      *
    
       n; i
    
      ++
    
      )
 {
 
    
      while
    
       (r 
    
      >=
    
       
    
      0
    
       
    
      &&
    
       f1[i] 
    
      +
    
       f2[r] 
    
      >
    
       
    
      0
    
      )
 r
    
      --
    
      ;
 
    
      if
    
       (r 
    
      <
    
       
    
      0
    
      )
 
    
      break
    
      ;
 
    
      int
    
       temp 
    
      =
    
       r;
 
    
      while
    
       (temp 
    
      >=
    
       
    
      0
    
       
    
      &&
    
       f1[i] 
    
      +
    
       f2[temp] 
    
      ==
    
       
    
      0
    
      )
 ans
    
      ++
    
      , temp
    
      --
    
      ;
 }
 printf(
    
      "
    
      %d\n
    
      "
    
      , ans);
 
    
      return
    
       
    
      0
    
      ;
}