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 = 0. 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 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 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

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
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).

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+5];
int B[maxn+5];
int C[maxn+5];
int D[maxn+5];
int n;
inline void Hash_clear(){
    for(int i=0;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=0;
    for(int i=0;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=0;
        scanf("%d",&n);
        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++){
                int x=A[i]+B[j];
                
                Hash2[h(x)].push_back(x);
                
        }
        for(int i=0;i<n;i++)
            for(int j=0;j<n;j++){
                int x=C[i]+D[j];
                cnt+=count_(-x);
            }
        cout<<cnt<<endl;
        if(T)
            cout<<endl;
    }
    //time__;
    return 0;
}