4 Values whose Sum is 0(POJ 2785)

• 原题如下：
  4 Values whose Sum is 0

  Time Limit: 15000MS		Memory Limit: 228000K
  Total Submissions: 29246		Accepted: 8887
  Case Time Limit: 5000MS

  Description

  The SUM problem can be formulated as follows: given four lists A, B, C, D of integer values, compute how many quadruplet (a, b, c, d ) ∈ A x B x C x D are such that a + b + c + d = 0 . In the following, we assume that all lists have the same size n .

  Input

  The first line of the input file contains the size of the lists n (this value can be as large as 4000). We then have n lines containing four integer values (with absolute value as large as 2²⁸ ) that belong respectively to A, B, C and D .

  Output

  For each input file, your program has to write the number quadruplets whose sum is zero.

  Sample Input

  6
  -45 22 42 -16
  -41 -27 56 30
  -36 53 -37 77
  -36 30 -75 -46
  26 -38 -10 62
  -32 -54 -6 45
  

  Sample Output

  5
  

  Hint

  Sample Explanation: Indeed, the sum of the five following quadruplets is zero: (-45, -27, 42, 30), (26, 30, -10, -46), (-32, 22, 56, -46),(-32, 30, -75, 77), (-32, -54, 56, 30).
• 题解：从4个数列中选择的话共有n⁴种情况，如果全都判断一遍肯定不可行，但如果将它们对半分成AB和CD再考虑，就可以快速解决了。从2个数列种选择的话只有n²种组合，所以可以进行枚举，先从A,B中取出a，b后，为了使总和为0则需要从C,D中取出c+d=-a-b，因此先将从C,D中取出数字的n²种方法全都枚举出来，将这些和排好序，这样就可以二分搜索了，复杂度O(n²logn).
• 代码：

  #include <cstdio>
  #include <cctype>
  #include <algorithm>
  #include <cmath>
  #include <cstring>
  #define number s-'0'
  
  using namespace std;
  
  const int MAX_N=5000;
  int A[MAX_N], B[MAX_N], C[MAX_N], D[MAX_N];
  int CD[MAX_N*MAX_N];
  int N;
  
  void read(int &x){
      char s;
      x=0;
      bool flag=0;
      while(!isdigit(s=getchar()))
          (s=='-')&&(flag=true);
      for(x=number;isdigit(s=getchar());x=x*10+number);
      (flag)&&(x=-x);
  }
  
  void write(int x)
  {
      if(x<0)
      {
          putchar('-');
          x=-x;
      }
      if(x>9)
          write(x/10);
      putchar(x%10+'0');
  }
  
  double calc(int);
  
  int main()
  {
      read(N);
      for (int i=0; i<N; i++)
      {
             read(A[i]);read(B[i]);read(C[i]);read(D[i]);
      }
      for (int i=0; i<N; i++)
          for (int j=0; j<N; j++)
              CD[i*N+j]=C[i]+D[j];
      sort(CD,CD+N*N);
      long long res=0;
      for (int i=0; i<N; i++)
          for (int j=0; j<N; j++)
          {
              int cd=-(A[i]+B[j]);
              res+=upper_bound(CD, CD+N*N, cd)-lower_bound(CD, CD+N*N, cd);
          }
      printf("%d\n", res);
  }