POJ 2785 4 Values whose Sum is 0

4 Values whose Sum is 0

Time Limit: 15000MS		Memory Limit: 228000K
Total Submissions: 20015		Accepted: 5974
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

 

 

 

解析：折半枚举。直接从4个数列中选择的话有n⁴种情况，时间复杂度太高，不可取。可以采取折半枚举的方式，将A、B、 C、 D分为AB和CD考虑，从A、B中取出a、b后，为了使总和为0则需要从C、D中取出c+d = -(a+b)。所以先将C、D取数的n²种方法枚举出来，并排好序，这样就可以运用二分搜索了，时间复杂度为O(n²logn²)。

 

 

 

#include <cstdio>
#include <algorithm>
#include <vector>
#define ll long long
using namespace std;

const int MAXN = 4000+5;
int a[MAXN], b[MAXN], c[MAXN], d[MAXN];
int cd[MAXN*MAXN];
int n;
pair<vector<int>::iterator, vector<int>::iterator> it;

void solve()
{
    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);
    ll res = 0;
    for(int i = 0; i < n; ++i){
        for(int j = 0; j < n; ++j){
            int c_d = -(a[i]+b[j]);
            it = equal_range(cd, cd+n*n, c_d);
            res += it.second-it.first;
        }
    }
    printf("%I64d\n", res);
}

int main()
{
    scanf("%d", &n);
    for(int i = 0; i < n; ++i)
        scanf("%d%d%d%d", &a[i], &b[i], &c[i], &d[i]);
    solve();
    return 0;
}