4 Values whose Sum is 0 POJ - 2785（二分查找）

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 2  ²⁸ ) 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).

题目大意：从四列中分别选一个数，问使所选的数加和为0，有多少种选法。

思路：暴力的话复杂度是n^4，肯定超时，用二分（个人认为叫折半枚举更合适）简化的话，复杂度有所降低。这里说下折半枚举，就是分别枚举计算第1、2行 和第3、4行的加和分别是sum1[i]、sum2[i]，再通过枚举sum1[i],sum2[i]计算结果。 然而这样仍然超时，这里在查找sum1，sum2时，我们是逐个枚举的，太耗时。下面介绍一个简化复杂度的技巧：利用二分查找。先对sum1、sum2排序，逐个枚举sum1里的数，在sum2中查找 存在几个sum2=（-sum1）。这里用到upper_bound和lower_bound查找。

 这里：利用 函数 upper_bound 和lower_bound 的性质  upper_bound(sum1,sum1+n*n,-sum2[i])-lower_bound(sum1,sum1+n*n,-sum2[i])  表示在sum1中查找值等于（-sum2）的个数。

下面上代码：

#include<stdio.h>
#include<iostream>
#include <map>
#include<set>
#include<algorithm>
#define inf 0x3f3f3f3f
using namespace std;
long long sum1[16100000],sum2[16100000],a[4100][4];
int main(){
    int n,i,j;
    long long sum=0;
    scanf("%d",&n);
    for(i=0;i<n;i++){
        scanf("%lld%lld%lld%lld",&a[i][0],&a[i][1],&a[i][2],&a[i][3]);
    }
    for(i=0;i<n;i++){
        for(j=0;j<n;j++){
            sum1[i*n+j]=a[i][0]+a[j][1];
            sum2[i*n+j]=a[i][2]+a[j][3];
        }
    }
    sort(sum1,sum1+n*n); //注意sum1、sum2的范围是n*n
    sort(sum2, sum2+n*n);
    for(i=0;i<n*n;i++){
        sum+=upper_bound(sum1,sum1+n*n,-sum2[i])-lower_bound(sum1,sum1+n*n,-sum2[i]);
        //在sum1中查找 值等于（-sum2） 的个数
    }
    printf("%lld\n",sum);
    return 0;
}