POJ - 2785 4 Values whose Sum is 0 —— 二分

4 Values whose Sum is 0

Time Limit: 15000MS		Memory Limit: 228000K
Total Submissions: 24643		Accepted: 7441
Case 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 -46
26 -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).

题意：给出四个序列，依次从中取一个数，问有多少种取法使得四个数的和为0。

思路：和取三个数的方式差不多，只要预处理其中任意两个序列就变成取三个数的问题，On2遍历其中两个序列，得出要找的数二分找最后的数。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <vector>
#include <cstring>
#include <cmath>
#include <stack>
#define ll long long
using namespace std;

vector<ll>v[5];
int n;
int main()
{
scanf("%d",&n);
int i,j;
ll cnt=0;
for(i=1;i<=n;i++)
{
ll a,b,c,d;
scanf("%lld%lld%lld%lld",&a,&b,&c,&d);
v[0].push_back(a);
v[1].push_back(b);
v[2].push_back(c);
v[3].push_back(d);
}
for(i=0;i<n;i++)
{
for(j=0;j<n;j++)
{
v[4].push_back(v[2][i]+v[3][j]);
}
}
sort(v[4].begin(),v[4].end());
for(i=0;i<n;i++)
{
for(j=0;j<n;j++)
{
ll k=v[0][i]+v[1][j];
k=k*-1;
cnt+=upper_bound(v[4].begin(),v[4].end(),k)-lower_bound(v[4].begin(),v[4].end(),k);
}
}
printf("%lld\n",cnt);
}