二分贪心 D

题目：

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

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).

分析：

在a,b,c,d四组中分别找出四个数相加，找出相加为0的个数简单的用回溯会超时，可以先把这四组变成两组，再用二分

代码：

#include<iostream> 
#include<iomanip> 
#include<algorithm> 
#include<stdio.h> 
using namespace std; 
long long int a[4005],b[4005],c[4005],d[4005],h[16000005],f[16000005]; 
int sum=0,n; 
void vp(int k) 
{ 
    int l,mid,r,x1,x2; 
    long long int x; 
    l=0; 
    r=n*n-1; 
    x=-1*h[k]; 
    while(l<=r) 
    { 
        mid=(l+r)/2; 
        if(f[mid]<x) 
            l=mid+1; 
        else 
        if(f[mid]>x) 
            r=mid-1; 
        else 
        { x1=x2=mid; 
            while(f[x1]==x&&x1>=0) 
            { 
                sum++; 
                x1--; 
            } 
            while(f[x2+1]==x&&x2+1<n*n) 
            { 
                sum++; 
                x2++; 
            } 
            r=l-1; 
        } 
    } 
} 
int main() 
{ 
    int i,k,j; 
    scanf("%d",&n); 
    for(i=0;i<n;i++) 
    scanf("%lld%lld%lld%lld",&a[i],&b[i],&c[i],&d[i]); 
    for(i=0;i<n;i++) 
        for(j=0;j<n;j++) 
        h[i*n+j]=a[i]+b[j]; 
    for(i=0;i<n;i++) 
        for(j=0;j<n;j++) 
        f[i*n+j]=c[i]+d[j]; 
    sort(h,h+n*n); 
    sort(f,f+n*n); 
    for(i=0;i<n*n;i++) 
        vp(i); 
    printf("%d\n",sum); 
}