BZOJ5317：[JSOI2018]战争(闵可夫斯基和)

令 (ain A,bin B) 则移动向量 (omega) 使得存在 (b+omega=a)
那么 (omega) 需要满足 (omega=a−b)
黑科技：闵可夫斯基和
直接构造闵可夫斯基和 (C={a+(−b)})
余下问题便是判断输入的移动向量是否在 (C) 内
可以强行使凸包的最下面为 ((0,0))，这样只要找到与坐标轴夹角最接近的边就好了

# include <bits/stdc++.h>
using namespace std;
typedef long long ll;

const int maxn(2e5 + 5);
const double eps(1e-13);
const double pi(acos(-1));
const double inf(1e15);

struct Point2D {
	double x, y;

	inline Point2D(double _x = 0, double _y = 0) {
		x = _x, y = _y;
	}

	inline Point2D operator +(Point2D ad) const {
		return Point2D(x + ad.x, y + ad.y);
	}

	inline Point2D operator -(Point2D ad) const {
		return Point2D(x - ad.x, y - ad.y);
	}

	inline double operator ^(Point2D ad) const { //dot
		return x * ad.x + y * ad.y;
	}

	inline double operator *(Point2D ad) const { //cross
		return x * ad.y - y * ad.x;
	}

	inline Point2D operator *(double ad) const {
		return Point2D(x * ad, y * ad);
	}

	inline double Len() {
		return sqrt(x * x + y * y);
	}

	inline double Angle() {
		return atan2(y, x);
	}
};

struct Segment2D {
	Point2D x, y;

	inline Segment2D(Point2D _x = Point2D(0, 0), Point2D _y = Point2D(0, 0)) {
		x = _x, y = _y;
	}
};

inline Point2D CrossPoint2D(Segment2D a, Segment2D b) {
	double k1, k2, t;
	k1 = (b.y - a.x) * (a.y - a.x);
	k2 = (a.y - a.x) * (b.x - a.x);
	t = k2 / (k1 + k2);
	return b.x + (b.y - b.x) * t;
}

Point2D tmp[maxn];

inline int Cmp(Point2D x, Point2D y) {
	return (x - tmp[1]) * (y - tmp[1]) > 0;
}

inline void Graham(Point2D *a, int &len) {
	int l = 0, mn = 0, i;
	for (i = 1; i <= len; ++i)
		if (!mn || (a[i].x < a[mn].x || (a[i].x == a[mn].x && a[i].y < a[mn].y))) mn = i;
	swap(a[1], a[mn]), tmp[l = 1] = a[1], sort(a + 2, a + len + 1, Cmp);
	for (i = 2; i <= len; ++i) {
		while (l > 1 && (a[i] - tmp[l - 1]) * (tmp[l] - tmp[l - 1]) >= 0) --l;
		tmp[++l] = a[i];
	}
	for (i = 1; i <= l; ++i) a[i] = tmp[i];
	len = l;
}

int n, m, q, len;
Point2D a[maxn], b[maxn], c[maxn], p;

inline void Minkowski() {
	int i, j;
	c[len = 1] = a[1] + b[1], a[0] = a[1], b[0] = b[1];
	for (i = 1; i < n; ++i) a[i] = a[i + 1] - a[i];
	for (i = 1; i < m; ++i) b[i] = b[i + 1] - b[i];
	a[n] = a[0] - a[n], b[m] = b[0] - b[m];
	for (i = j = 1; i <= n || j <= m; )
		if (j > m || (i <= n && a[i] * b[j] >= 0)) ++len, c[len] = c[len - 1] + a[i++];
		else ++len, c[len] = c[len - 1] + b[j++];
}

inline int Query() {
	int l, r, mid, cur = 2;
	p = p - c[1];
	if (c[len] * p > 0 || p * c[2] > 0) return 0;
	l = 2, r = len - 1;
	while (l <= r) {
		mid = (l + r) >> 1;
		if (c[mid] * p > 0) l = mid + 1, cur = mid;
		else r = mid - 1;
	}
	return (p - c[cur]) * (c[cur + 1] - c[cur]) <= 0;
}

int main() {
	int i;
	scanf("%d%d%d", &n, &m, &q);
	for (i = 1; i <= n; ++i) scanf("%lf%lf", &a[i].x, &a[i].y);
	for (i = 1; i <= m; ++i) scanf("%lf%lf", &b[i].x, &b[i].y), b[i] = b[i] * -1;
	Graham(a, n), Graham(b, m), Minkowski(), Graham(c, len);
	for (i = 2; i <= len; ++i) c[i] = c[i] - c[1];
	for (i = 1; i <= q; ++i) scanf("%lf%lf", &p.x, &p.y), printf("%d
", Query());
	return 0;
}