一眼n<=18狀壓dp……方程什麽的都很顯然，枚舉兩只小鳥，再將這條拋物線上的小鳥抓出來就好啦。只是這樣O(n^3)的dp必然是要TLE的，我一開始這樣交上去顯然跑得巨慢無比，後來轉念一想：面對一個嶄新的情況的時候，只有搭配的優劣之分，沒有先後的區別，所以最外面的一層可以直接去掉，變成O(n^2)的dp。這樣就跑的很快啦~

PS:print()函數只是調試輸出，作用是輸出now 的二進制形式+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 = 0, k = 1;
    char c;
    c = getchar();
    while(c < ‘0‘ || c > ‘9‘) { if(c == ‘-‘) k = -1; c = getchar(); }
    while(c >= ‘0‘ && c <= ‘9‘
 
) x = x * 10 + c - ‘0‘, 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[30], tot = 0;
    int k = now;
    while(k) 
    {
        a[++ tot] = k & 1;
        k >>= 1;
    }
    cout << now << " ";
    for(int i = 1; i <= tot; i ++) cout << a[i];
    for(int i = n; i > tot; i --) cout <<‘0‘;
    cout << " " << dp[now];
    cout << endl;
}

void DP(int now)
{
    if(dp[now] != INF) return;
    for(int i = 0; i < n; i ++)
    {
        if(((1 << i) & now)) continue;
        for(int j = i + 1; j < n; j ++)
        {
            if(x[i] == x[j]) continue;
            Get_ab(x[i], y[i], x[j], y[j]);
            if(a >= 0) continue;
            int aft = 0;
            for(int k = 0; k < n; k ++)
                if(On_Line(x[k], y[k])) aft = (aft | (1 << k));
            int tem = aft | now;
            DP(tem);
            dp[now] = min(dp[now], dp[tem] + 1);
        }
        int aft = (1 << i);
        int tem = aft | now;
        DP(tem);
        dp[now] = min(dp[now], dp[tem] + 1);
        break;
    }
}

void init()
{
    len = (1 << n) - 1;
    for(int i = 0; i < len; i ++) dp[i] = INF;
}

int main()
{
    T = read();
    while(T --)
    {
        n = read(), m = read();
        init();
        for(int i = 0; i < n; i ++)
            scanf("%lf%lf", &x[i], &y[i]);
        dp[len] = 0;
        DP(0);
        printf("%d\n", dp[0]);
    }
    return 0;
}

【題解】NOIP2016憤怒的小鳥