之前做完飛揚的小鳥的時候還以為這道題和它一樣。。

題意：用最少的鳥把豬打完。
題解：看n的範圍可知是狀壓DP。列舉每兩隻豬之間的拋物線，若拋物線符合題意就看這條拋物線還覆蓋了哪些豬。而有些豬可能不會和任意一隻豬在同一拋物線上，所以還要新增一條拋物線只覆蓋一隻豬。
狀態轉移方程：dp[i|para[j]]=min(dp[i|para[j]],dp[i]+1)$dp[i|para[j]]=min(dp[i|para[j]],dp[i]+1)$

#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
 

using namespace std;
const double eps = 1e-8;
const int MAXN = 21;

struct Node{
    double x, y;
}P[MAXN];

int para[MAXN * MAXN], dp[(1 << 18) + 5];

inline void Solve(double x1, double y1, double x2, double y2, double &a, double &b){
    a = (y2 * x1 - y1 * x2) / (x1 * x2 * x2 - x1 * x1 * x2);
    b = (y2 * x1 * x1 - y1 * x2 * x2) / (x1 * x1 * x2 - x1 * x2 * x2);
}

inline
 
 bool check(double a, double b, double x, double y){ //檢查第li條拋物線與第pi只豬是否相連 
    return fabs(a * x * x + b * x - y) < eps;
} 

int main(){
    freopen("in.txt", "r", stdin);
    int T; scanf("%d", &T);
    while(T--){
        int n, m, cnt = 0; scanf("%d%d", &n, &m);
        for(int i = 1
 
; i <= n; i++)
            scanf("%lf%lf", &P[i].x, &P[i].y);
        for(int i = 1; i <= n; i++){
            para[++cnt] = (1 << (i - 1)); //這條拋物線只經過一隻小豬 
            for(int j = 1, vis = 0; j < i; j++){
                if(fabs(P[i].x - P[j].x) < eps) continue;
                if(vis & (1 << (j - 1))) continue; //之前已經有拋物線經過這隻豬了 
                double a, b;

                Solve(P[i].x, P[i].y, P[j].x, P[j].y, a, b);
                if(a >= 0) continue; //不合法 
                para[++cnt] = (1 << (i - 1)); //這條拋物線起初只經過一隻小豬 
                for(int k = 1; k < i; k++){
                    if(check(a, b, P[k].x, P[k].y)){
                        vis |= (1 << (k - 1));
                        para[cnt] |= (1 << (k - 1));
                    }
                }
            }
        }
        memset(dp, 0x3f, sizeof(dp)); dp[0] = 0;
        for(int i = 0; i < (1 << n); i++){
            for(int j = 1; j <= cnt; j++){
                dp[i | para[j]] = min(dp[i | para[j]], dp[i] + 1);
            }
        }
        printf("%d\n", dp[(1 << n) - 1]);
    }
    return 0;
}

迴圈的時候把起點設為0會更好寫，但我不想改了233