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 << "]" << "\n";
     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, , BUF, stdin);
     return *bg++;
 } 

 inline int ri(){
     int x = , f = , c = gc();
     for(; c<||c>; f = c=='-'?-:f, c=gc());
     for(; c>&&c<; x = x* + c - , 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-;
 const int inf = 1e9 + ;
 const LL mod = 1e9 + ;
 const int maxN = 1e5 + ;
 const LL ONE = ;

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

 struct Matrix{
     double m[][];

     Matrix(){}
     Matrix(double x11, double x12, double x21, double x22) {
         m[][] = x11;
         m[][] = x12;
         m[][] = x21;
         m[][] = x22;
     }

     double det() {
         return m[][] * m[][] - m[][] * m[][];
     }
 };

 int T, n, m, ans;
 PDD pig[];
 // 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[ << ]; 

 int bitcount32(int bits) {
     bits = (bits & 0x55555555) + (bits >>  & 0x55555555);
     bits = (bits & 0x33333333) + (bits >>  & 0x33333333);
     bits = (bits & 0x0f0f0f0f) + (bits >>  & 0x0f0f0f0f);
     bits = (bits & 0x00ff00ff) + (bits >>  & 0x00ff00ff);
     return (bits & 0x0000ffff) + (bits >>  & 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()) == ) 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) >= ) 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) == ) return true;
     return false;
 }

 void solve() {
     int ret = ;

     // 枚举所有点对
     For(i, , n) {
         For(j, i + , 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 = ;
             st |=  << (i - );
             st |=  << (j - );
             For(k, , n) {
                 if(k == i || k == j) continue;
                 if(check(pig[k], p)) st |=  << (k - );
             }
             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( << 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[] = ;
     int len = state.size();

     Rep(i,  << 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] + );
         }
     }

     ans = f[( << n) - ];
 }

 int main(){
     INIT();
     cin >> T;
     while(T--) {
         state.clear();
         si.clear();
         cin >> n >> m;
         For(i, , n) cin >> pig[i];
         solve();

         cout << ans << endl;
     }

     return ;
 }