poj 2785 4 Values whose Sum is 0(hash)

4 Values whose Sum is 0

Time Limit: 15000MS		Memory Limit: 228000K
Total Submissions: 12631		Accepted: 3546
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

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). 

思路；hash，这里用开放定址法

AC代码;

#include <cstring>
#include <string>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <cmath>
#include <vector>
#include <cstdlib>
#include <iostream>

using namespace std;
const int maxn=19999997;
struct node
{
    int v,cnt;
}num[maxn];
int a[4005],b[4005],c[4005],d[4005];
void hash(int x)
{
    int loc=x%maxn;
    if(loc<0) loc+=maxn;
    while(num[loc].cnt&&num[loc].v!=x) loc=(loc+1)%maxn;
    num[loc].v=x;
    num[loc].cnt++;
}
int find(int x)
{
    int loc=x%maxn;
    if(loc<0) loc+=maxn;
    while(num[loc].cnt&&num[loc].v!=x) loc=(loc+1)%maxn;
    return num[loc].cnt;
}
int main()
{
    int n;
    scanf("%d",&n);
    for(int i=0;i<n;i++)
    scanf("%d%d%d%d",&a[i],&b[i],&c[i],&d[i]);
    memset(num,0,sizeof(num));
    for(int i=0;i<n;i++)
    for(int j=0;j<n;j++)
    {
        hash(a[i]+b[j]);
    }
    int ans=0;
    for(int i=0;i<n;i++)
    for(int j=0;j<n;j++)
    {
        ans+=find(-c[i]-d[j]);
    }
    cout<<ans<<endl;
    return 0;
}