暴力求调

看了题意后，不难写出如下暴力做法：枚举全排列，然后依次连线，没连一条线和前面线的不同交点个数就是当期新分出的块数。因为是圆环，所以答案减去 $1$。

#include <cstdio>
#include <vector>
#include <algorithm>
#include <cmath>
const double eps = 1e-8;
int n, a[25], ans, cnt;
struct twt {
	double x, y;
	twt(double a = 0, double b = 0) { x = a, y = b; }
	twt operator - (twt b) { return twt(x-b.x, y-b.y); }
	twt operator / (double b) { return twt(x/b, y/b); }
	twt operator + (twt b) { return twt(x+b.x, y+b.y); }
	twt operator * (double b) { return twt(x*b, y*b); }
	double operator * (twt b) { return x*b.x + y*b.y; }
	double operator ^ (twt b) { return x*b.y - b.x*y; }
	double len() { return sqrt(x*x + y*y); }
	bool operator == (twt b) {
		return fabs(x-b.x) <= eps && fabs(y-b.y) <= eps;
	}
	bool operator < (twt b) const {
		return x < b.x || (x == b.x && y < b.y);
	} 
	twt roa() { return twt(y, -x); }
};
std::vector<twt> s;
struct dwd {
	twt a, b;
	dwd(twt x = twt(), twt y = twt()) { a = x, b = y; }
} line[25];
twt inter(dwd a, dwd b) {
	twt p1 = a.a, v1 = a.b, p2 = b.a, v2 = b.b; 
	return p2 + (v2-p2) * ((v1-p1)^(p2-p1)) / ((v2-p2)^(v1-p1));
}		
int solve() {
	int an = 1;
	for (int i = 1; i <= n; i++) {
		dwd now = dwd(twt(i, 1), twt(a[i], 0));
		s.clear();
		for (int j = 1; j <= cnt; j++)
			if (line[j].b.x > a[i]) s.push_back(inter(now, line[j]));
		std::sort(s.begin(), s.end());
		s.erase(std::unique(s.begin(), s.end()), s.end());		
		line[++cnt] = now;
		an += s.size() + 1;
	}
	return an-1;
}
int main() {
	scanf("%d", &n);
	for (int i = 1; i <= n; i++) a[i] = i;
	do {
		cnt = 0;
		ans = std::max(ans, solve());
	} while(std::next_permutation(a+1, a+n+1));
	printf("%d", ans);
}

这份代码看上去没有任何的问题，然而在 $n=8$ 时答案是 $31$, 而不是标程的 $33$，所以 这题是错题 能告诉我哪里写错了吗。