POJ 2785 4 Values whose Sum is 0(折半枚举)

4 Values whose Sum is 0

Time Limit: 15000MS		Memory Limit: 228000K
Total Submissions: 18767		Accepted: 5571
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 30 -75 -46
  
  
  
  
  
  
  
  

   
   
   
   -36 53 -37 77
  
  
  
  
  
  
  
  

   
   
   
   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).

题意：有四个数列A，B，C，D。 每个数列n个数，在每个数列中都任意取出一个数。要求这四个数相加等于0，问这样的四个数组合共有多少组？

题解：直接暴力做的话O(n^4)，时间复杂度很大。我们来对半分成AB和CD来考虑。从两个队列中取数就是n^2个组合。 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C，D中取出数字的n^2种组合全部枚举出来，将这些和排序，这样就可以利用二分搜索找出满足的组合数。

代码如下:

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define maxn 4040
int a[maxn],b[maxn],c[maxn],d[maxn],cd[maxn*maxn];
int main()
{
	int n,i,j;
	while(scanf("%d",&n)!=EOF)
	{
		for(i=0;i<n;++i)//注意输入，竖向的 
			scanf("%d%d%d%d",&a[i],&b[i],&c[i],&d[i]);
		for(i=0;i<n;++i)
		{
			for(j=0;j<n;++j)
				cd[n*i+j]=c[i]+d[j];
		}
		sort(cd,cd+n*n);
		int ans=0;
		for(i=0;i<n;++i)
		{
			for(j=0;j<n;++j)
			{
				int k=-(a[i]+b[j]);
				ans+=upper_bound(cd,cd+n*n,k)-lower_bound(cd,cd+n*n,k);
			}//长度为n*n的有序数组中的k的个数。 
		}
		printf("%d\n",ans);
	}
	return 0;
}