ACM-二分-POJ-2785-4 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 -4626 -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).

主要用到了二分的思想，让我们计算四个的和并不好计算，但是我们可以二分成两个两个的和，分而治之。

然后对两个的和进行排序，让大的和小的(负数)相加，计算结果。代码如下:

/**POJ2785 ------ 2015/4/16*/#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <algorithm>#include <cstdlib>using namespace std;const int max_n = 5000;int a[max_n];int b[max_n];int c[max_n];int d[max_n];int pre[max_n*max_n];//用来记录前一部分的和int next1[max_n*max_n];//用来记录后一部分的和int n;/**返回A是否大于B 按降序排列*/bool com_max(int a,int b){    return a>b;}int main(){    //freopen("POJ2785.txt","r",stdin);    scanf("%d",&n);    for(int i=0;i<n;i++)    {        scanf("%d %d %d %d",a+i,b+i,c+i,d+i);    }    int index = 0;    for(int i=0;i<n;i++)    {        for(int j=0;j<n;j++)        {            pre[index] = a[i] + b[j];            next1[index] = c[i] + d[j];            index++;        }    }    //sort(pre,pre+index,com_max);    //sort(next1,next1+index,com_max);    sort(pre,pre+index);    sort(next1,next1+index);    int r = index-1;    int res = 0;    for(int i=0;i<index;i++)    {        /**pre是从大到小开始,next是从小到大开始*/        while(r>=0 && pre[i]+next1[r]>0)        {            //说明在i不变的情况下，next太小，得增加            r--;        }        if(r<0)        {            //查询完成            break;        }        /**        if((pre[i]+next1[r]) == 0)        {            res++;            r--;        }        */        /**一个i可能对应多个解*/        int temp = r;        while(temp >=0 && (pre[i]+next1[temp])==0)        {            res++;            temp--;        }    }    printf("%d\n",res);}