二分贪心-D

• 原题

  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
  5

• 思路&解析
  从每一列里各取一个数，求有多少组四个数的和加起来为零。
  先分别求出前两列后两列的所有的和，然后对后两列的和进行排序，对后两列进行二分查找看有多少种情况即可。

• AC代码

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<algorithm>
using namespace std;
int a[4002][4],sum1[16000002],sum2[16000002];
int main()
{
    int n,mid;
    while(~scanf("%d",&n))
    {
        for(int i=0;i<n; i++)
        {
            scanf("%d%d%d%d",&a[i][0],&a[i][1], &a[i][2],&a[i][3]);
        }
        int k=0;
        int m=0;
        for(int i=0;i<n;i++)
            for(int j=0;j<n;j++)
            {
                sum1[k++]=a[i][0]+a[j][1];
                sum2[m++]=a[i][2]+a[j][3];
            }
        sort(sum2,sum2+k);
        int cnt=0;
        for(int i=0;i<k;i++)
        {
            int left=0;
            int right=k-1;
            while(left<=right)
            {
                mid=(left+right)/2;
                if(sum1[i]+sum2[mid]==0)
                {
                    cnt++;
                    for(int j=mid+1;j<k;j++)
                    {
                        if(sum1[i]+sum2[j]!=0)
                           break;
                        else
                        cnt++;
                    }
                    for(int j=mid-1;j>=0;j--)
                    {
                        if(sum1[i]+sum2[j]!=0)
                           break;
                        else
                           cnt++;
                    }
                    break;
                }
                if(sum1[i]+sum2[mid]<0)
                    left=mid+1;
                else
                    right=mid-1;
            }
        }
        printf("%d\n",cnt);
    }
    return 0;
}