poj2785双向搜索

Posted 2020-09-13 walfy

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

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：求四个数的和为0的情况有几种

题解：折半枚举+sort，，很重要的小技巧：upper_bound(a,a+n,s)-low_bound（a,a+n,s）表示a数组（已排序）里等于s的个数，

刚开始用if(all[lower_bound(all,all+n*n,-a[i]-b[j])-all]==-a[i]-b[j])处理wa了，发现原来是因为只算了一个相等的情况，要是有几个同时等于就漏了

#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<cstdio>
#include<iomanip>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#define pi acos(-1)
#define ll long long
#define mod 1000000007

using namespace std;

const double g=10.0,eps=1e-9;
const int N=4000+5,maxn=10000+5,inf=0x3f3f3f3f;

ll a[N],b[N],c[N],d[N];
ll all[N*N];

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
 //   cout<<setiosflags(ios::fixed)<<setprecision(2);
    ll n;
    while(cin>>n){
        for(ll i=0;i<n;i++)cin>>a[i]>>b[i]>>c[i]>>d[i];
        for(ll i=0;i<n;i++)
        {
            for(ll j=0;j<n;j++)
            {
                all[i*n+j]=c[i]+d[j];
            }
        }
        sort(all,all+n*n);
        ll ans=0;
        for(ll i=0;i<n;i++)
        {
            for(ll j=0;j<n;j++)
            {
                ans+=upper_bound(all,all+n*n,-a[i]-b[j])-lower_bound(all,all+n*n,-a[i]-b[j]);
            }
        }
        cout<<ans<<endl;
    }
    return 0;
}

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