UVa 1152 -4 Values whose Sum is 0—[哈希表实现]

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 × B × C × D are such that a + b + c + d = . In the following, we
assume that all lists have the same size n.
Input
The input begins with a single positive integer on a line by itself indicating the number of the cases
following, each of them as described below. This line is followed by a blank line, and there is also a
blank line between two consecutive inputs.
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 test case, your program has to write the number quadruplets whose sum is zero.
The outputs of two consecutive cases will be separated by a blank line.
Sample Input
1

-   -
- -
-  -
-  - -
 - -
- - - 45

Sample Output

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

解题思路：

　　枚举并存储A＋B的和，然后枚举C＋D，搜索－C－D的个数，问题的关键是如何存储A+B的和。本题数据量不小，极限数据n＝4000时，A＋B的和有16，000，000个，数组显然开不下。那么不妨建立哈希表来存储。

代码如下：

 #include <iostream>
 #include <cstdio>
 #include <cstdlib>
 #include <cstring>
 #include <set>
 #include <vector>
 #include <ctime>
 #define H 1000000
 #define maxn 4000
 #define time__ cout<<" time: "<<double(clock())/CLOCKS_PER_SEC<<endl;
 using namespace std;
 vector<int> Hash2[H];

 int A[maxn+];
 int B[maxn+];
 int C[maxn+];
 int D[maxn+];
 int n;
 inline void Hash_clear(){
     for(int i=;i<H;i++)
         Hash2[i].clear();
 }
 inline int h(int x){
     return abs(x%H);
 }
 inline int count_(int x){

     int h_=h(x);
     int cnt=;
     for(int i=;i<Hash2[h_].size();i++)
         if(Hash2[h_][i]==x) cnt++;
     return cnt;

 }
 int main(int argc, const char * argv[]) {

     int T;
     scanf("%d",&T);
     while (T--) {
         Hash_clear();

         int cnt=;
         scanf("%d",&n);
         for(int i=;i<n;i++)
             scanf("%d%d%d%d",&A[i],&B[i],&C[i],&D[i]);
         for(int i=;i<n;i++)
             for(int j=;j<n;j++){
                 int x=A[i]+B[j];

                 Hash2[h(x)].push_back(x);

         }
         for(int i=;i<n;i++)
             for(int j=;j<n;j++){
                 int x=C[i]+D[j];
                 cnt+=count_(-x);
             }
         cout<<cnt<<endl;
         if(T)
             cout<<endl;
     }
     //time__;
     return ;
 }