4 Values whose Sum is 0

这个题目同上道二分的题目一样，只是把数字分成两堆，在排序用二分的思想，但是必须明白，不是在找到一个的条件的情况下跳出，可能只有重复的。这里是重点。继续我的二分之路，我的时间复杂度很高，效率不怎么好。看了下别人的代码，好像都是用hash，下一站就是hash了。

Time Limit: 15000MS Memory Limit: 228000K
Total Submissions: 11586 Accepted: 3249
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

代码：
[cpp]
<span style="font-family:FangSong_GB2312;font-size:18px;">#include<iostream>
#include<cmath>
#include<vector>
#include<algorithm>
#define maxn 4004
using namespace std;
int map1[maxn*maxn];
int map2[maxn*maxn];
int a[maxn],b[maxn],c[maxn],d[maxn];
int main()
{
      int n,i,j,k,sum,p;
      scanf("%d",&n);
      for(i=0;i<n;i++)
      {
         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++)
         map1[i*n+j]=a[i]+b[j];
      for(i=0;i<n;i++)
        for(j=0;j<n;j++)
          map2[i*n+j]=c[i]+d[j];
      sort(map1,map1+n*n);
      sort(map2,map2+n*n);
      sum=0;
      p=n*n-1;
      for(i=0;i<n*n;i++)
      {
        while(p>=0&&map1[i]+map2[p]>0) p--;
        if(p<0) break;
        int temp=p;
        while(temp>=0&&map1[i]+map2[temp]==0)
        {
            sum++; temp--;
        }
      }
      printf("%d\n",sum);
      //system("pause");
       return 0;
}

</span>

补充：软件开发 , C++ ,