4 Values whose Sum is 0

Time Limit: 15000MS		Memory Limit: 228000K
Total Submissions: 19243		Accepted: 5744
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 2

²⁸ ) 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 -4626 -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).

 

题目大意：四个数列，问你有几种可能的组合满足，a[i]+b[j]+c[k]+d[e]==0;

思路分析：直接遍历所有情况的话，复杂度是n^4。肯定超时，本题应该采用二分做，先进行数组合并，将四个数组合成为两个n*n的数组，然后对

其中一个排序，这样就可以用二分查找另一个数组是否存在对应的数，这样复杂度就变成了n*nlogn^2，这样就不会超时了。

代码：

#include <iostream>

#include <algorithm>

#include <cstdio>

#include <cstring>

#include <queue>

#include <stack>

#define mem(a) memset(a,0,sizeof(a))

using namespace std;

const int inf=0xffffff;

const int maxn=4000+10;

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

int f1[maxn*maxn],f2[maxn*maxn];

int main()

{

    int n;

    while(scanf("%d",&n)!=EOF)

    {

        int cut=0;

        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];

                f2[i*n+j]=c[i]+d[j];

            }

        }

        sort(f2,f2+n*n);

    for(int i=0;i<n*n;i++)

    {

        int low=0,high=n*n-1;

    while(low<=high)

    {

        int mid=(low+high)/2;

        if(f2[mid]==-f1[i])

        {

             cut++;

             for(int j=mid-1;j>=0;j--)

             {

                 if(f2[j]!=-f1[i]) break;

                 cut++;

             }

             for(int j=mid+1;j<n*n;j++)

             {

                  if(f2[j]!=-f1[i]) break;

                 cut++;

             }

             break;

        }

      else if(f2[mid]<-f1[i]) low=mid+1;

      else high=mid-1;

    }

    }

    cout<<cut<<endl;

    }

    return 0;

}