# POJ 2785：4 Values whose Sum is 0 （双向BFS）

4 Values whose Sum is 0

 Time Limit: 15000MS Memory Limit: 228000K Total Submissions: 20020 Accepted: 5977 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).

AC代码：

#include<cstdio>
#include<math.h>
#include<iostream>
#include<algorithm>
#define max(a,b) (a>b?a:b)
using namespace std;
__int64 a[4005],b[4005],c[4005],d[4005];
__int64 cd[4005*4005];
int main()
{
int n;
scanf("%d",&n);
for(int i=0; i<n; i++)
scanf("%I64d%I64d%I64d%I64d",a+i,b+i,c+i,d+i);
for(int i=0; i<n; i++)
for(int j=0; j<n; j++)
cd[i*n+j]=c[i]+d[j];
sort(cd,cd+n*n);
__int64 ans=0;
for(int i=0; i<n; i++)
for(int j=0; j<n; j++)
{
__int64 ab=-(a[i]+b[j]);
ans+=upper_bound(cd,cd+n*n,ab)-lower_bound(cd,cd+n*n,ab);
}
printf("%I64d\n",ans);
return 0;
}