poj2785 简单二分

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 .

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 .

For each input file, your program has to write the number quadruplets whose sum is zero.

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

5

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：四个数列，问你有几种可能的组合满足，a[i]+b[j]+c[k]+d[e]==0;

思路分析：直接遍历所有情况的话，复杂度是n^4。肯定超时，本题应该采用二分做，先进行数组合并，将四个数组合成为两个n*n的数组，然后对

其中一个排序，这样就可以用二分查找另一个数组是否存在对应的数，这样复杂度就变成了n*nlogn^2，这样就不会超时了。

代码：

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <queue>
#include <stack>
#define mem(a) memset(a,0,sizeof(a))
using namespace std;
const int inf=0xffffff;
const int maxn=4000+10;
int a[maxn],b[maxn],c[maxn],d[maxn];
int f1[maxn*maxn],f2[maxn*maxn];
int main()
{
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        int cut=0;
        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++)
            {
                f1[i*n+j]=a[i]+b[j];
                f2[i*n+j]=c[i]+d[j];
            }
        }
        sort(f2,f2+n*n);
    for(int i=0;i<n*n;i++)
    {
        int low=0,high=n*n-1;
    while(low<=high)
    {
        int mid=(low+high)/2;
        if(f2[mid]==-f1[i])
        {
             cut++;
             for(int j=mid-1;j>=0;j--)
             {
                 if(f2[j]!=-f1[i]) break;
                 cut++;
             }
             for(int j=mid+1;j<n*n;j++)
             {
                  if(f2[j]!=-f1[i]) break;
                 cut++;
             }
             break;
        }
      else if(f2[mid]<-f1[i]) low=mid+1;
      else high=mid-1;
    }
    }
    cout<<cut<<endl;
    }
    return 0;
}