AC但疑惑（数据过水？）

#include<bits/stdc++.h>
#define x first
#define y second
using namespace std;

typedef long long ll;
typedef pair<ll, ll> pll;
typedef pair<double, double> pdd;

const ll INF = 1e18;
const double EPS = 1e-8;

pdd Computing_Function(pdd pig1, pdd pig2) {    //计算pig1和pig2和(0, 0)的二次函数系数
    double a = 
        (pig2.x*pig1.y - pig1.x*pig2.y) / ( (pig1.x*pig1.x)*pig2.x - pig1.x*(pig2.x*pig2.x) ), 
           b = 
        ((pig2.x*pig2.x)*pig1.y - (pig1.x*pig1.x)*pig2.y) / ( pig1.x*(pig2.x*pig2.x)-(pig1.x*pig1.x)*pig2.x );
    // cout << a << " " << b << "\n";
    return pdd{a, b};
}

bool Determine_Track(double a, double b, pdd pig) {    //判断pig是否在给定的抛物线上
    return fabs((a * pig.x * pig.x + b * pig.x) - pig.y) < EPS;
}

void solve() {
    ll n, m;
    cin >> n >> m;
    vector<pdd> pigs(n + 5);
    unordered_set<ll> track;     // 存储一个二进制表示这个二进制数所表示的两个小猪可以被一起消灭
    vector<ll> dp(300000, INF);    // dp[s]表示打掉小猪🐷的状态为s时最少需要的小鸟数量
    
    for(int i = 1; i <= n; i++) cin >> pigs[i].x >> pigs[i].y;
    for(int i = 1; i < n; i++) {
        for(int j = i + 1; j <= n; j++) {
            double a = Computing_Function(pigs[i], pigs[j]).first, b = Computing_Function(pigs[i], pigs[j]).second;
            if(a < 0) {
                ll index = ((1ll << (i - 1)) | (1ll << (j - 1)));
                for(int k = 1; k <= n; k++) {
                    if(k == i || k == j) continue;
                    if(Determine_Track(a, b, pigs[k])) index |= (1ll << (k - 1));
                }
                track.insert(index);
            }
        }
    }
        
    dp[0] = 0;

    for(ll s = 0; s <= (1 << n) - 1; s++) {
        for(auto x : track) {
            dp[s | x] = min(dp[s | x], dp[s] + 1);
        }
        for(int i = 1; i <= n; i++) {
            if(((1ll << (i - 1)) & s) == 0) {
                dp[s | (1ll << (i - 1))] = min(dp[s | (1ll << (i - 1))], dp[s] + 1);
            }
        }
    }

    cout << dp[(1 << n) - 1] << "\n";
}

int main() {
    ll t;
    cin >> t;
    while(t--) solve();
    return 0;
}

#include<bits/stdc++.h>
#define x first
#define y second
using namespace std;

typedef long long ll;
typedef pair<ll, ll> pll;
typedef pair<double, double> pdd;

const ll INF = 1e18;
const double EPS = 1e-8;

pdd Computing_Function(pdd pig1, pdd pig2) {    //计算pig1和pig2和(0, 0)的二次函数系数
    /*解方程
    {
    pig1.y = a * pig1.x * pig1.x + b * pig1.x, 
    pig2.y = a * pig2.x * pig2.x + b * pig2.x
    }，
    容易得：*/

    // double a = 
    //     (pig2.x*pig1.y - pig1.x*pig2.y) / ( (pig1.x*pig1.x)*pig2.x - pig1.x*(pig2.x*pig2.x) ), 
    //        b = 
    //     ((pig2.x*pig2.x)*pig1.y - (pig1.x*pig1.x)*pig2.y) / ( pig1.x*(pig2.x*pig2.x)-(pig1.x*pig1.x)*pig2.x );
    // 弃用，可能存在其他情况未处理
    
    double x1 = pig1.x, y1 = pig1.y;
    double x2 = pig2.x, y2 = pig2.y;
    
    if(fabs(x1) < EPS || fabs(x2) < EPS) return pdd{0, 0};

    if(fabs(x1 - x2) < EPS) return pdd{0, 0};

    double ad = ( (x1*x1)*x2 - x1*(x2*x2) ), bd = ( x1*(x2*x2)-(x1*x1)*x2 );

    if(fabs(ad) < EPS) return pdd{0, 0};
    if(fabs(bd) < EPS) return pdd{0, 0};
    
    double a = (x2*y1 - x1*y2) / ad, b = ((x2*x2)*y1 - (x1*x1)*y2) / bd;
    
    return pdd{a, b};
}

bool Determine_Track(double a, double b, pdd pig) {    //判断pig是否在给定的抛物线上
    return fabs((a * pig.x * pig.x + b * pig.x) - pig.y) < EPS;
}

void solve() {
    ll n, m;
    cin >> n >> m;
    vector<pdd> pigs(n + 5);
    unordered_set<ll> track;     // 存储一个二进制表示这个二进制数所表示的两个小猪可以被一起消灭
    vector<ll> dp(300000, INF);    // dp[s]表示打掉小猪🐷的状态为s时最少需要的小鸟数量
    
    for(int i = 1; i <= n; i++) cin >> pigs[i].x >> pigs[i].y;
    for(int i = 1; i < n; i++) {
        for(int j = i + 1; j <= n; j++) {
            pdd x = Computing_Function(pigs[i], pigs[j]);
            double a = x.first, b = x.second;
            if(a == 0 && b == 0) continue;
            if(a < EPS) {
                ll index = ((1ll << (i - 1)) | (1ll << (j - 1)));
                for(int k = 1; k <= n; k++) {
                    if(k == i || k == j) continue;
                    if(Determine_Track(a, b, pigs[k])) index |= (1ll << (k - 1));
                }
                track.insert(index);
            }
        }
    }

    dp[0] = 0;

    for(ll s = 0; s <= (1 << n) - 1; s++) {
        for(auto x : track) {
            dp[s | x] = min(dp[s | x], dp[s] + 1);
        }
        for(int i = 1; i <= n; i++) {
            if(((1ll << (i - 1)) & s) == 0) {
                dp[s | (1ll << (i - 1))] = min(dp[s | (1ll << (i - 1))], dp[s] + 1);
            }
        }
    }

    cout << dp[(1 << n) - 1] << "\n";
}

int main() {
    ll t;
    cin >> t;
    while(t--) solve();
    return 0;
}