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 =  . 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 ). We then have n lines containing four integer values (with absolute value as large as  ) 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

-   -
- -
-  -
-  - -
 - -
- - - 

Sample Output

Hint

Sample Explanation: Indeed, the sum of the five following quadruplets is zero: (-, -, , ), (, , -, -), (-, , , -),(-, , -, ), (-, -, , ).

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=;i<n;i++){
        for(int j=;j<n;j++){
           CD[i*n+j]=C[i]+D[j];
        }
    }
    sort(CD,CD+n*n);
    ll ans=;
    for(int i=;i<n;i++){
        for(int j=;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\n",ans);
 }
 int main()
 {
     while(scanf("%d",&n)==){
        for(int i=;i<n;i++){
           for(int j=;j<;j++){
              scanf("%d",&mp[i][j]);
           }
        }
        for(int i=;i<n;i++){
           A[i] = mp[i][];
           B[i] = mp[i][];
           C[i] = mp[i][];
           D[i] = mp[i][];
        }
        solve();
     }
     return ;
 }