POJ 2785 4 Values whose Sum is 0(二分)

4 Values whose Sum is 0

Time Limit: 15000MS Memory Limit: 228000KTotal Submissions: 14089 Accepted: 3975Case Time Limit: 5000MS

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 2²⁸ ) 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

跟以前一个题目很相似，不过那个有五堆，枚举一二堆合并堆，三四堆合并堆，然后再二分第五堆。这个直接将前两堆合并为1堆，后两堆合并为一堆，枚举第一堆，二分第二堆。

   分析时间复杂度，O(4000*4000*log2(4000*4000))，不过这个是最优的时候，二分的代码写得并不好，找到边界之后需要遍历，然后依次再移动，假如合并之后a和b全为0,那时间复杂度就远远不止这点了。变为了O（10^14）........数据比较水。每次二分找的时候可以加个标记，这样时间复杂度会降下来，自己就没实现这个了。。空间也给了很大，几千万的数组是妥妥可以放的。

   题目地址：4 Values whose Sum is 0

AC代码：

#include<iostream>#include<cstdio>#include<algorithm>using namespace std;int a[16000005];    //第一堆int b[16000005];    //第二堆int p[4005][4];int t;int erfen(int x){    int cnt=0;    int l=0,r=t-1,mid;    while(r>l)    {        mid=(l+r)>>1;        if(b[mid]>=x) r=mid;        else l=mid+1;    }    while(b[l]==x&&l<t)    //依次左移    {        cnt++;        l++;    }    return cnt;}int main(){    int n,i,j;    long long res;    while(cin>>n)    {        res=0;        for(i=0;i<n;i++)            for(j=0;j<4;j++)                scanf("%d",&p[i][j]);        t=0;        //合并前两个数作为第一堆，后两个数作为第二堆        for(i=0;i<n;i++)            for(j=0;j<n;j++)                a[t++]=p[i][0]+p[j][1];        sort(a,a+t);        t=0;        for(i=0;i<n;i++)            for(j=0;j<n;j++)                b[t++]=p[i][2]+p[j][3];        sort(b,b+t);        for(i=0;i<t;i++)            res+=erfen(-a[i]);        cout<<res<<endl;    }    return 0;}/*6-45 22 42 -16-41 -27 56 30-36 53 -37 77-36 30 -75 -4626 -38 -10 62-32 -54 -6 45*///49300K5282MS