一眼n<=18状压dp……方程什么的都很显然，枚举两只小鸟，再将这条抛物线上的小鸟抓出来就好啦。只是这样O(n^3)的dp必然是要TLE的，我一开始这样交上去显然跑得巨慢无比，后来转念一想：面对一个崭新的情况的时候，只有搭配的优劣之分，没有先后的区别，所以最外面的一层可以直接去掉，变成O(n^2)的dp。这样就跑的很快啦~
PS:print()函数只是调试输出，作用是输出now 的二进制形式+dp[now]；
#include <bits/stdc++.h>
using namespace std;
#define db double
#define eps 0.00000001
#define maxn 30
#define maxm (1 << 18) + 20
#define INF 999999
int T, n, m, dp[maxm], len;
db a, b, x[maxn], y[maxn]; int read()
{
int x = , k = ;
char c;
c = getchar();
while(c < '' || c > '') { if(c == '-') k = -; c = getchar(); }
while(c >= '' && c <= '') x = x * + c - '', c = getchar();
return x * k;
} void Get_ab(db x1, db y1, db x2, db y2)
{
a = (y1 * x2 - y2 * x1) / (x1 * x1 * x2 - x2 * x2 * x1);
b = (x2 * x2 * y1 - x1 * x1 * y2) / (x1 * x2 * x2 - x1 * x1 * x2);
} bool On_Line(db x1, db y1)
{
db r = x1 * x1 * a + x1 * b;
if((r - y1 < eps) && (r - y1 > -eps)) return true;
else return false;
} void print(int now)
{
int a[], tot = ;
int k = now;
while(k)
{
a[++ tot] = k & ;
k >>= ;
}
cout << now << " ";
for(int i = ; i <= tot; i ++) cout << a[i];
for(int i = n; i > tot; i --) cout <<'';
cout << " " << dp[now];
cout << endl;
} void DP(int now)
{
if(dp[now] != INF) return;
for(int i = ; i < n; i ++)
{
if((( << i) & now)) continue;
for(int j = i + ; j < n; j ++)
{
if(x[i] == x[j]) continue;
Get_ab(x[i], y[i], x[j], y[j]);
if(a >= ) continue;
int aft = ;
for(int k = ; k < n; k ++)
if(On_Line(x[k], y[k])) aft = (aft | ( << k));
int tem = aft | now;
DP(tem);
dp[now] = min(dp[now], dp[tem] + );
}
int aft = ( << i);
int tem = aft | now;
DP(tem);
dp[now] = min(dp[now], dp[tem] + );
break;
}
} void init()
{
len = ( << n) - ;
for(int i = ; i < len; i ++) dp[i] = INF;
} int main()
{
T = read();
while(T --)
{
n = read(), m = read();
init();
for(int i = ; i < n; i ++)
scanf("%lf%lf", &x[i], &y[i]);
dp[len] = ;
DP();
printf("%d\n", dp[]);
}
return ;
}