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).
题意：求在四组数中各挑选一个满足a+b+c+d==0的个数。
直接使用BFS时间复杂度是O（n^4），显然会超时，我们可以把整个四组数据分成两组，然后就是对两组数据分别BFS，最终统计结果，这样便把时间复杂度降低到了O（n^2），很容易就可以过啦~
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;
}
