poj 2785 4 Values whose Sum is 0（折半枚举（双向搜索））

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).

AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <set>
using namespace std;
#define ll long long
#define N 4006
int n;
int mp[N][N];
int A[N],B[N],C[N],D[N];
int CD[N*N];
void solve(){
   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);
   ll ans=0;
   for(int i=0;i<n;i++){
       for(int j=0;j<n;j++){
          int cd = -(A[i]+B[j]);
          ans+=upper_bound(CD,CD+n*n,cd)-lower_bound(CD,CD+n*n,cd);
       }
   }
   printf("%I64d
",ans);
}
int main()
{
    while(scanf("%d",&n)==1){
       for(int i=0;i<n;i++){
          for(int j=0;j<4;j++){
             scanf("%d",&mp[i][j]);
          }
       }
       for(int i=0;i<n;i++){
          A[i] = mp[i][0];
          B[i] = mp[i][1];
          C[i] = mp[i][2];
          D[i] = mp[i][3];
       }
       solve();
    }
    return 0;
}