哈希-4 Values whose Sum is 0

4 Values whose Sum is 0
Time Limit: 15000MS Memory Limit: 228000K
Total Submissions: 17875 Accepted: 5255
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).

Source
Southwestern Europe 2005
哈希的一道比较简单的题,
题意:给你四个集合,从四个集合中分别选出一个元素,使四个元素的和为零,问有几种选法;
方法:先让两个集合加和,用哈希链表,储存计算的结果,有后两个集合的计算结果查找,不过直接写链表可能比较耗时,我交了一次11s多,后来换成前向星3s多.

#include <map>#include <set>#include <list>#include <cmath>#include <queue>#include <stack>#include <vector>#include <string>#include <cstdio>#include <cstring>#include <cstdlib>#include <iostream>#include <algorithm>using namespace std;#define eps 1e-9#define LL long long#define PI acos(-1.0)#define INF 0x3f3f3f3f#define CRR fclose(stdin)#define CWW fclose(stdout)#define RR freopen("input.txt","r",stdin)#define WW freopen("output.txt","w",stdout)const int MAX = 4010;const int Mod = 10000007;struct node{    int num;    int data;    int next;} H[Mod*10];int head[Mod];int a[MAX],b[MAX],c[MAX],d[MAX];int top;int Creat(int ans){    H[top].num=1;    H[top].data=ans;    return top++;}int main(){    int n;    int ans;    int p,q;    int Max;    while(~scanf("%d",&n))    {        memset(head,-1,sizeof(head));        for(int i=0; i<n; i++)        {            scanf("%d %d %d %d",&a[i],&b[i],&c[i],&d[i]);        }        top=0;        for(int i=0; i<n; i++)        {            for(int j=0; j<n; j++)            {                ans=a[i]+b[j];                p=abs(ans)%Mod;                q=head[p];                while(q!=-1)                {                    if(H[q].data==ans)                    {                        H[q].num++;                        break;                    }                    q=H[q].next;                }                if(q==-1)                {                    q=Creat(ans);                    H[q].next=head[p];                    head[p]=q;                }            }        }        Max=0;        for(int i=0; i<n; i++)        {            for(int j=0; j<n; j++)            {                ans=-1*(c[i]+d[j]);                p=abs(ans)%Mod;                q=head[p];                while(q!=-1)                {                    if(H[q].data==ans)                    {                        Max+=H[q].num;                        break;                    }                    q=H[q].next;                }            }        }        printf("%d\n",Max);    }    return 0;}