NOIP2016提高组复赛C 愤怒的小鸟

题目链接：http://uoj.ac/problem/265

题目大意：

　　太长了不想概括。。。

分析：

　　状压DP的模板题，把所有可能的抛物线用二进制表示，然后暴力枚举所有组合，详情见代码内注释

代码如下：

#pragma GCC optimize("Ofast")
#include <bits/stdc++.h>
using namespace std;

#define INIT() std::ios::sync_with_stdio(false);std::cin.tie(0);
#define Rep(i,n) for (int i = 0; i < (n); ++i)
#define For(i,s,t) for (int i = (s); i <= (t); ++i)
#define rFor(i,t,s) for (int i = (t); i >= (s); --i)
#define foreach(i,c) for (__typeof(c.begin()) i = c.begin(); i != c.end(); ++i)
#define rforeach(i,c) for (__typeof(c.rbegin()) i = c.rbegin(); i != c.rend(); ++i)

#define pr(x) cout << #x << " = " << x << "  "
#define prln(x) cout << #x << " = " << x << endl

#define LOWBIT(x) ((x)&(-x))

#define ALL(x) x.begin(),x.end()
#define INS(x) inserter(x,x.begin())

#define ms0(a) memset(a,0,sizeof(a))
#define msI(a) memset(a,inf,sizeof(a))
#define msM(a) memset(a,-1,sizeof(a))

#define pii pair<int,int> 
#define piii pair<pair<int,int>,int> 
#define MP make_pair
#define PB push_back
#define ft first
#define sd second

template<typename T1, typename T2>
istream &operator>>(istream &in, pair<T1, T2> &p) {
    in >> p.first >> p.second;
    return in;
}

template<typename T>
istream &operator>>(istream &in, vector<T> &v) {
    for (auto &x: v)
        in >> x;
    return in;
}

template<typename T1, typename T2>
ostream &operator<<(ostream &out, const std::pair<T1, T2> &p) {
    out << "[" << p.first << ", " << p.second << "]" << "
";
    return out;
}

inline int gc(){
    static const int BUF = 1e7;
    static char buf[BUF], *bg = buf + BUF, *ed = bg;
    
    if(bg == ed) fread(bg = buf, 1, BUF, stdin);
    return *bg++;
} 

inline int ri(){
    int x = 0, f = 1, c = gc();
    for(; c<48||c>57; f = c=='-'?-1:f, c=gc());
    for(; c>47&&c<58; x = x*10 + c - 48, c=gc());
    return x*f;
}

typedef long long LL;
typedef unsigned long long uLL;
typedef pair< double, double > PDD;
typedef set< int > SI;
typedef vector< int > VI;
const double EPS = 1e-10;
const int inf = 1e9 + 9;
const LL mod = 1e9 + 7;
const int maxN = 1e5 + 7;
const LL ONE = 1;

int sgn(double x) {
    if(fabs(x) < EPS) return 0;
    return x > 0 ? 1 : -1;
}

struct Matrix{
    double m[3][3];
    
    Matrix(){}
    Matrix(double x11, double x12, double x21, double x22) {
        m[1][1] = x11;
        m[1][2] = x12;
        m[2][1] = x21;
        m[2][2] = x22;
    }
    
    double det() {
        return m[1][1] * m[2][2] - m[1][2] * m[2][1];
    }
};

int T, n, m, ans;
PDD pig[20];
// 用n位二进制数记录一条抛物线可能穿过的点的状态 
// 比如抛物线过1号，5号，7号点，那么数值为 ：000000000001010001 
VI state; 
SI si;
// f[i]为当前状态的最小步数 
int f[1 << 18]; 

int bitcount32(int bits) {
    bits = (bits & 0x55555555) + (bits >> 1 & 0x55555555);
    bits = (bits & 0x33333333) + (bits >> 2 & 0x33333333);
    bits = (bits & 0x0f0f0f0f) + (bits >> 4 & 0x0f0f0f0f);
    bits = (bits & 0x00ff00ff) + (bits >> 8 & 0x00ff00ff);
    return (bits & 0x0000ffff) + (bits >> 16 & 0x0000ffff);
}

// 通过2个点算抛物线参数[a, b]
bool calcAB(PDD x, PDD y, PDD &para) {
    Matrix D = Matrix(x.ft * x.ft, x.ft, y.ft * y.ft, y.ft);
    if(sgn(D.det()) == 0) return false;
    Matrix D1 = Matrix(x.sd, x.ft, y.sd, y.ft);
    Matrix D2 = Matrix(x.ft * x.ft, x.sd, y.ft * y.ft, y.sd);
    
    para.ft = D1.det() / D.det();
    if(sgn(para.ft) >= 0) return false; 
    para.sd = D2.det() / D.det();
    return true;
}

// 验证点x是否符合抛物线
bool check(PDD x, PDD &para) {
    if(sgn(para.ft * x.ft * x.ft + para.sd * x.ft - x.sd) == 0) return true;
    return false;
}

void solve() {
    int ret = 0;
    
    // 枚举所有点对 
    For(i, 1, n) {
        For(j, i + 1, n) {
            PDD p;
            if(!calcAB(pig[i], pig[j], p)) continue;
            // 看是否有其他点也经过这条抛物线
            int st = 0;
            st |= 1 << (i - 1);
            st |= 1 << (j - 1);
            For(k, 1, n) {
                if(k == i || k == j) continue;
                if(check(pig[k], p)) st |= 1 << (k - 1);
            }
            if(si.find(st) == si.end()) {
                si.insert(st);
                state.PB(st);
            }
        }
    }
    // 把只过一个点的抛物线也存一下
    Rep(i, n) state.PB(1 << i);
    // 把所有抛物线都得到后，问题就变成在这些抛物线中最少能选取几条，进行或运算后二进制1~n位全为1 
    msI(f);
    f[0] = 0;
    int len = state.size();
    
    Rep(i, 1 << n) {
        // 当枚举到f[i]时，f[i]已经是最优解了 
        Rep(j, len) {
            f[i | state[j]] = min(f[i | state[j]], f[i] + 1);
        }
    }
    
    ans = f[(1 << n) - 1];
}

int main(){
    INIT(); 
    cin >> T;
    while(T--) {
        state.clear();
        si.clear();
        cin >> n >> m;
        For(i, 1, n) cin >> pig[i];
        solve();
        
        cout << ans << endl;
    }
    
    return 0;
}