POJ - 2785 4 Values whose Sum is 0(二分枚举)

最新推荐文章于 2022-03-20 19:33:27 发布

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最新推荐文章于 2022-03-20 19:33:27 发布

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本文链接：https://blog.youkuaiyun.com/sUb_569/article/details/54428350

本文介绍了一个经典的编程问题——SUM问题的解决方法。该问题要求在四个整数列表中找到所有四元组，使得这些四元组的元素之和等于零，并统计这样的四元组数量。文章提供了完整的算法实现及样例解释。

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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).
题意
a,b,c,d四个数组，从每一个数组中挑一个形成一个组合，求最多有多少个组合数为0.

#include<iostream>
#include<algorithm>
using namespace std;
#define N 4002
int a[N],b[N],c[N],d[N],e[N*N],f[N*N];
int main(){
    std::ios_base::sync_with_stdio(false);
    int t,i,j,q=0;
    cin>>t;

    for(i=0;i<t;i++)    
        cin>>a[i]>>b[i]>>c[i]>>d[i];
    for(i=0;i<t;i++)
        for(j=0;j<t;j++)
            e[i*t+j]=a[i]+b[j];
    for(i=0;i<t;i++)
        for(j=0;j<t;j++)
            f[i*t+j]=c[i]+d[j];
    sort(e,e+t*t);
    sort(f,f+t*t);  
    //把f数组中的数在e数组中查找，upper_bound-lower_bound返回值为1，表示能找到 
    for(i=0;i<t*t;i++)
            q+=upper_bound(e,e+t*t,-f[i])-lower_bound(e,e+t*t,-f[i]);
    cout<<q;    

}