文章标题POJ 2785：4 Values whose Sum is 0？（二分）

最新推荐文章于 2020-09-29 23:35:44 发布

Wang_SF2015

最新推荐文章于 2020-09-29 23:35:44 发布

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本文介绍了一种解决四列表求和为零问题的有效算法。通过对列表进行分组、枚举并结合二分查找的方法，高效地找到所有符合条件的四元组。此算法巧妙利用了upper_bound和lower_bound函数来减少不必要的操作。

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4 Values whose Sum is 0

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 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

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：题目要我们从四个表中各找一个数相加的和为0。
分析：直接暴力破解肯定不行，可先对四个数表分成两组进行枚举相加，然后其中一个枚举，另一个二分，记住，记录重复的情况。
代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<math.h>
#include<algorithm>
using namespace std;
int a[4005],b[4005],c[4005],d[4005];
int sum1[4005*4005],sum2[4005*4005];
int bs (int a[],int key,int k){
    int lo=0,hi=k-1;
    int mi,ans=hi+1;
    while (lo<=hi){
        mi=((hi-lo)>>1)+lo;
        //if (a[mi]==key)return mi;
        if (a[mi]<key)lo=mi+1;
        else {
            hi=mi-1;ans=min(ans,mi);    
        }
    }
    if (ans==hi+1)
        return -1;
    return ans;
}
int main ()
{
    int n;
    while (scanf ("%d",&n)!=EOF){
        for (int i=0;i<n;i++){
            scanf ("%d%d%d%d",&a[i],&b[i],&c[i],&d[i]);
        }
        int k1=0;
        for (int i=0;i<n;i++){
            for (int j=0;j<n;j++){
                sum1[k1++]=a[i]+b[j];
            }
        }
        int k2=0;
        for (int i=0;i<n;i++){
            for (int j=0;j<n;j++){
                sum2[k2++]=c[i]+d[j];
            }
        }
        sort (sum1,sum1+k1);
        int count=0;
        for (int i=0;i<k2;i++){
                count+=upper_bound(sum1,sum1+k1,-sum2[i])-lower_bound(sum1,sum1+k1,-sum2[i]);//点睛之笔 
        }
        printf ("%d\n",count);
    }           
    return 0;
}