POJ2785 4 Values whose Sum is 0（折半枚举）

题目链接

4 Values whose Sum is 0

Time Limit: 15000MS		Memory Limit: 228000K
Total Submissions: 29564		Accepted: 9001
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).

 

给定各有n个整数的四个数列A，B，C，D。要从每个数列中各取出一个数，使四个数的和为0。求出这样的组合的个数。当一个数列中有多个相同的数字时，把它们作为不同的数字看待。

 

如果直接四层循环做的话复杂度是，显然不可行。

如果将它们对半分成AB和CD再考虑，就可以快速解决了。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中取数字的种组合全部枚举出来，排好序后就能运用二分搜索了。总复杂度为。

 

AC代码：

#include<iostream>
#include<sstream>
#include<algorithm>
#include<string>
#include<cstring>
#include<iomanip>
#include<vector>
#include<cmath>
#include<ctime>
#include<stack>
#include<queue>
#define e 2.71828182
#define Pi 3.141592654
using namespace std;
int CD[4000*4000];
int main()
{
    int n,A[4000],B[4000],C[4000],D[4000];
    cin>>n;
    for(int i=0;i<n;i++) cin>>A[i]>>B[i]>>C[i]>>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 ans=0;
    for(int i=0;i<n;i++)
        {
            for(int j=0;j<n;j++)
            {
            int AB=-(A[i]+B[j]);
            //取出C和D中和为AB的部分
            ans+=upper_bound(CD,CD+n*n,AB)-lower_bound(CD,CD+n*n,AB); 
        }
    }
    cout<<ans;
}