Solution

数据范围识状压
挺好想的状压dp
对于每只猪考虑两种情况：自己被单独打下来或者被其他抛物线经过打下来
现在已经有一个确定点 ((0 ,0)) 再加一只猪，两个点无法确定一条抛物线
也就是说这条抛物线是可以给你自己规划的
也也就是说自己被单独打下来一定可行
不过能打多一点会更优
三点确定抛物线
我们暴力枚举两只没被打下来的猪
看还能打几只
更新dp即可

注意两点：
预处理出所有抛物线能打几只猪，可以省一维枚举每只猪，不然 (TLE)
此题卡精度，判断两个浮点数是否相等可以确定一个精度（此题设置为 (10^{-6})）

Code

#include<iostream>
#include<cstdio>
#include<queue>
#include<cstring>
#include<algorithm>
#include<climits>
#include<cmath>
#define LL long long
#define REP(i, x, y) for(int i = (x);i <= (y);i++)
using namespace std;
int RD(){
    int out = 0,flag = 1;char c = getchar();
    while(c < '0' || c >'9'){if(c == '-')flag = -1;c = getchar();}
    while(c >= '0' && c <= '9'){out = out * 10 + c - '0';c = getchar();}
    return flag * out;
    }
const int maxn = (1 << 19);
int T;
int num;
int dp[maxn], one[25];
double a, b;
inline void build(double &a,double &b,double x1,double y1,double x2,double y2){
    a=(x2*y1-x1*y2)/(x1*x2*(x1-x2));
    b=(x1*x1*y2-x2*x2*y1)/(x1*x2*(x1-x2));
	}//计算a,b
bool judge(double x, double y){
	double y1 = a * x * x + b * x;
	if(fabs(y - y1) <= 1e-6)return 1;
	return 0;
	}
struct Node{
	double x, y;
	}I[25];
int van[25][25];
void init(){
	memset(van, 0, sizeof(van));
	REP(i, 1, num){
		REP(j, i + 1, num){
			if(fabs(I[i].x - I[j].x) <= 1e-6)continue;
			build(a, b, I[i].x, I[i].y, I[j].x, I[j].y);
			if(a >= 0)continue;
			REP(k, 1, num){
				if(judge(I[k].x, I[k].y))van[i][j] |= one[k];
				}
			}
		}
	}
int main(){
	T = RD();
	while(T--){
		num = RD(), RD();
		REP(i, 1, num)scanf("%lf %lf", &I[i].x, &I[i].y), one[i] = (1 << (i - 1));
		int maxstate = (1 << num) - 1;
		REP(i, 0, maxstate)dp[i] = 1e9;
		init();
		dp[0] = 0;
		REP(i, 0, maxstate){//状态
			REP(j, 1, num){//一只新猪
				if(i & one[j])continue;
				dp[i | one[j]] = min(dp[i | one[j]], dp[i] + 1);//看看只打一只猪赚不赚
				REP(k, j + 1, num){//另一只新猪
					if(i & one[k] || !van[j][k])continue;
					dp[i | van[j][k]] = min(dp[i | van[j][k]], dp[i] + 1);
					}
				}
			}
		printf("%d
", dp[maxstate]);
		}
	return 0;
	}