【题解】NOIP2016愤怒的小鸟

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 ;
}