POJ 2785 - 4 Values whose Sum is 0（折半枚举）

4 Values whose Sum is 0

Time Limit: 15000MS	Memory Limit: 228000K
Case Time Limit: 5000MS

Description

The SUM problemcan 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 aresuch that a + b + c + d = 0 . In the following, we assume that all lists havethe same size n .

Input

The first line ofthe input file contains the size of the lists n (this value can be as large as4000). We then have n lines containing four integer values (with absolute valueas large as 2²⁸ ) that belong respectively to A, B, C and D .

Output

For each inputfile, 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

SampleExplanation: 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】

给定各有n个整数的四个数列A,B,C,D。要从每个数列中各取出一个数，使得四个数字和为零。求出这样的组合的个数。当一个数列中有多个相同的数字时，把它们当作不同的数字看待。

1<=n<=4000

|序列中的数字|<=2^28

 

【思路】

直接用四重循环枚举复杂度是O(n^4)不可行。所以这里用到了折半枚举的思想，先枚举前面两个数组的和的所有可能项，并存入数组中，排序，那么再去逐个枚举后面两个数组的和的可能项，假如从后两个数组中取出来c,d,判断数组中是否有一项x==-(c+d)即可，可以利用二分搜索使复杂度降为O(n^2*logn)。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;

typedef long long ll;
const int maxn = 4050;

int n;
int a[maxn], b[maxn], c[maxn], d[maxn];
int flip[maxn*maxn];

int main() {
	while (scanf("%d", &n) == 1) {
		for (int i = 0; i < n; i++) {
			scanf("%d%d%d%d", &a[i], &b[i], &c[i], &d[i]);
		}
		for (int i = 0; i < n; i++) {
			for (int j = 0; j < n; j++)
				flip[i * n + j] = a[i] + b[j];
		}
		sort(flip, flip + n*n);

		ll ans = 0;
		for (int i = 0; i < n; i++) {
			for (int j = 0; j < n; j++) {
				int sum = -(c[i] + d[j]);
				ans += upper_bound(flip, flip + n*n, sum) - lower_bound(flip, 
flip + n*n, sum);
			}
		}
		printf("%lld\n", ans);
	}
	return 0;
}