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;}