洛谷 P2831 愤怒的小鸟

思路

未优化

状压$\text{DP}$

$n\le 18$，不是暴搜就是状压，因为我$jio$得状压会比较好理解，所以就写一篇状压的题解叭

首先我们要预处理出经过任意两点的抛物线可以击中的小猪有哪些，可以用$line[i][j]$来表示经过$i,j$的抛物线经过的小猪的集合，集合用二进制数来表示

• 这里有一个小问题就是如何求抛物线$y=a{x}^2+bx$中的$a,b$

  假设目前的抛物线经过$(x_1,y_1)$和$(x_2,y_2)$两点，已知$x>0$，那么有

  $$y_1=a{x}_1^2+b{x}_1$$

  $$y_2=a{x}_2^2+b{x}_2$$

  则

  $$a{x}_1+b=\frac{y_1}{x_1}$$

  $$a{x}_2+b=\frac{y_2}{x_2}$$

  两式做差得

  $$a(x_1-x_2)=\frac{y_1}{x_1}-\frac{y_2}{x_2}$$

  所以

  $a=\frac{\frac{y_1}{x_1}-\frac{y_2}{x_2}}{(x_1-x_2)}$

  $b=\frac{y_1}{x_1}-a\ast x_1$

处理完之后就要想一想如何$\text{DP}$

我们设$dp[s]$表示消灭集合$s$内所有小猪所用的最少的小鸟数

显然$dp[0]=0$，因为没有猪当然用不到鸟

假设当前的状态为$s$，抛物线为经过$i,j$点的抛物线，这条抛物线打掉的小猪的状态为$line[i][j]$，那么有

$$dp[s|line[i][j]]=\min (dp[s|line[i][j]],dp[s]+1)$$

其中$s|line[i][j]$表示当前状态$s$在增加了经过$i,j$点的这条抛物线之后能打到的小猪的集合，显然要从$dp[s|line[i][j]]$和$dp[s]+1$中取最小

时间复杂度$O(T{n}^2{2}^n)$O(能过才怪)，在洛谷上吸氧($O2$)可以过

优化

随意选择$s$内的一只小猪$j$，那么$j$最后一定会被一只小鸟消灭，所以我们固定住这只小猪$j$，只枚举$k$转移

更详细见AThousandSuns的题解

时间复杂度$O(Tn{2}^n)$，稳了

代码

未优化

/*
Author:loceaner
*/
#include 
#include 
#include 
#include 
#include 
#define eps (1e-6)
using namespace std;

const int A = 5e5 + 11;
const int B = 1e6 + 11;
const int mod = 1e9 + 7;
const int inf = 0x3f3f3f3f;

inline int read() {
	char c = getchar(); int x = 0, f = 1;
	for( ; !isdigit(c); c = getchar()) if(c == '-') f = -1;
	for( ; isdigit(c); c = getchar()) x = x * 10 + (c ^ 48);
	return x * f;
}

int n, m, num, line[20][20], dp[B];
struct Pig { double x, y; } p[A]; //猪猪结构体 

bool cmp(Pig a, Pig b) { //根据坐标排序猪猪 
	return (a.x != b.x) ? (a.x < b.x) : (a.y < b.y);
} 

struct Line { //抛物线结构体，包含ax^2+bx中的a和b 
	double a, b;
	bool judge(Pig qwq) { //判断一个点是否在这个抛物线上 
		double val = a * qwq.x * qwq.x + b * qwq.x, Y = qwq.y;
		return ((val - Y) >= -eps && (val - Y) <= eps); 
	}
};

inline Line get(Pig a, Pig b) { //已知两点坐标，求这两点所处的抛物线
	//初中知识，推导见思路 
	Line ans;
	ans.a = (a.y / a.x - b.y / b.x) / (a.x - b.x);
	ans.b = (a.y / a.x - ans.a * a.x);
	return ans;	
}

int main() {
	int T = read();
	while (T--) {
		n = read(), m = read();
		memset(line, 0, sizeof(line)); //多测不清空，爆零两行泪 
		memset(dp, 0x3f, sizeof(dp));
		for (int i = 1; i <= n; i++) scanf("%lf%lf", &p[i].x, &p[i].y);
		sort(p + 1, p + 1 + n, cmp); //排序猪猪 
		for (int i = 1; i <= n; i++) line[i][i] = 1 << (i - 1); 
		//处理只经过一个点的抛物线，不加就会人没了…… 
		for (int i = 1; i <= n; i++) //预处理出所有抛物线 
			for (int j = 1; j < i; j++) {
				Line now = get(p[i], p[j]);
				if (now.a >= 0) continue;
				for (int k = 1; k <= n; k++) //line[i][j]抛物线能经过的猪猪集合 
					if (now.judge(p[k])) line[i][j] |= 1 << (k - 1);
				line[j][i] = line[i][j];
			}
		int U = (1 << n) - 1; //全集 
		dp[0] = 0; //没猪当然是0 
		for (int i = 1; i <= n; i++) //快乐地DP吧！ 
			for (int j = 1; j <= n; j++) 
				for (int k = 0; k <= U; k++) 
					dp[k | line[i][j]] = min(dp[k | line[i][j]], dp[k] + 1);
		cout << dp[U] << '\n';
	} 
	return 0;
}

优化

/*
Author:loceaner
*/
#include 
#include 
#include 
#include 
#include 
#define eps (1e-6)
using namespace std;

const int A = 5e5 + 11;
const int B = 1e6 + 11;
const int mod = 1e9 + 7;
const int inf = 0x3f3f3f3f;

inline int read() {
	char c = getchar(); int x = 0, f = 1;
	for( ; !isdigit(c); c = getchar()) if(c == '-') f = -1;
	for( ; isdigit(c); c = getchar()) x = x * 10 + (c ^ 48);
	return x * f;
}

int n, m, num, line[20][20], dp[B], must[B];
struct Pig { double x, y; } p[A];

bool cmp(Pig a, Pig b) {
	return (a.x != b.x) ? (a.x < b.x) : (a.y < b.y);
} 

struct Line {
	double a, b;
	bool judge(Pig qwq) {
		double val = a * qwq.x * qwq.x + b * qwq.x, Y = qwq.y;
		return ((val - Y) >= -eps && (val - Y) <= eps); 
	}
};

inline Line get(Pig a, Pig b) {
	Line ans;
	ans.a = (a.y / a.x - b.y / b.x) / (a.x - b.x);
	ans.b = (a.y / a.x - ans.a * a.x);
	return ans;	
}

int main() {
	for (int i = 0; i < (1 << 18); i++) {
		int j = 1;
		for (; j <= 18 && i & (1 << (j - 1)); j++);
		must[i] = j;
	}
	int T = read();
	while (T--) {
		num = 0;
		n = read(), m = read();
		memset(line, 0, sizeof(line)); //多测不清空，爆零两行泪 
		memset(dp, 0x3f, sizeof(dp));
		for (int i = 1; i <= n; i++) scanf("%lf%lf", &p[i].x, &p[i].y);
		sort(p + 1, p + 1 + n, cmp);
//		for (int i = 1; i <= n; i++) line[i][i] = 1 << (i - 1);
		//因为下面枚举must[i]的时候就是枚举的这个，所以就不需要了 
		for (int i = 1; i <= n; i++) 
			for (int j = 1; j < i; j++) {
				Line now = get(p[i], p[j]);
				if (now.a >= 0) continue;
				for (int k = 1; k <= n; k++) 
					if (now.judge(p[k])) line[i][j] |= 1 << (k - 1);
				line[j][i] = line[i][j];
			}
		int U = (1 << n) - 1;
		dp[0] = 0;
		for (int i = 0; i <= U; i++) {
			int j = must[i];
			dp[i | (1 << (j - 1))] = min(dp[i | (1 << (j - 1))], dp[i] + 1); //单独处理 
			for (int k = 1; k <= n; k++) 
				dp[i | line[j][k]] = min(dp[i | line[j][k]], dp[i] + 1);
		}
		cout << dp[U] << '\n';
	} 
	return 0;
}