UVA10256 The Great Divide

洛谷传送门

UVA 传送门

考虑对两个点集求出凸包，显然如果这两个凸包相离就合法，然后问题就转化成了这两个凸包是否有交。

设红点凸包包围的点集为 \(A\)，蓝点凸包包围的点集为 \(B\)，问题为询问是否 \(\exists a \in A,b \in B,a=b\)，即 \(a-b=0\)。

于是对所有蓝点取反，求出新的蓝点凸包，再与红点凸包计算它们闵可夫斯基和的凸包，如果 \((0,0)\) 在它的内部（包含边界），就说明 \(A\) 和 \(B\) 有交，输出 No；否则输出 Yes。

时间复杂度 \(O((n + m) \log (n + m))\)，瓶颈在于对点排序。

code

const int maxn = 510;

int n, m, stk[maxn], top;
struct node {
	int x, y;
	node(int a = 0, int b = 0) : x(a), y(b) {}
};

node operator + (node a, node b) {
	return node(a.x + b.x, a.y + b.y);
}

node operator - (node a, node b) {
	return node(a.x - b.x, a.y - b.y);
}

int operator * (node a, node b) {
	return a.x * b.y - a.y * b.x;
}

bool cmp(node a, node b) {
	return a.x < b.x || (a.x == b.x && a.y < b.y);
}

vector<node> convex(vector<node> vc) {
	sort(vc.begin(), vc.end(), cmp);
	top = 0;
	int n = (int)vc.size();
	if (n == 1) {
		return vc;
	}
	for (int i = 0; i < n; ++i) {
		while (top >= 2 && (vc[stk[top]] - vc[stk[top - 1]]) * (vc[i] - vc[stk[top - 1]]) <= 0) {
			--top;
		}
		stk[++top] = i;
	}
	vector<node> res;
	for (int i = 1; i < top; ++i) {
		res.pb(vc[stk[i]]);
	}
	top = 0;
	for (int i = n - 1; ~i; --i) {
		while (top >= 2 && (vc[stk[top]] - vc[stk[top - 1]]) * (vc[i] - vc[stk[top - 1]]) <= 0) {
			--top;
		}
		stk[++top] = i;
	}
	for (int i = 1; i < top; ++i) {
		res.pb(vc[stk[i]]);
	}
	return res;
}

vector<node> mkfsj(vector<node> a, vector<node> b) {
	int n = (int)a.size(), m = (int)b.size();
	vector<node> va(n), vb(m), res;
	res.pb(a[0] + b[0]);
	for (int i = 0; i < n; ++i) {
		va[i] = a[(i + 1) % n] - a[i];
	}
	for (int i = 0; i < m; ++i) {
		vb[i] = b[(i + 1) % m] - b[i];
	}
	int i = 0, j = 0;
	while (i < n && j < m) {
		res.pb(res.back() + (va[i] * vb[j] >= 0 ? va[i++] : vb[j++]));
	}
	while (i < n) {
		res.pb(res.back() + va[i++]);
	}
	while (j < m) {
		res.pb(res.back() + vb[j++]);
	}
	res.erase(res.begin());
	return res;
}

bool check(vector<node> a, vector<node> b) {
	vector<node> res = mkfsj(a, b);
	node O;
	int n = (int)res.size();
	for (int i = 1; i < n; ++i) {
		if ((O - res[i - 1]) * (res[i] - res[i - 1]) > 0) {
			return 0;
		}
	}
	return 1;
}

void solve() {
	vector<node> a, b;
	for (int i = 0, x, y; i < n; ++i) {
		scanf("%d%d", &x, &y);
		a.pb(node(x, y));
	}
	for (int i = 0, x, y; i < m; ++i) {
		scanf("%d%d", &x, &y);
		b.pb(node(-x, -y));
	}
	vector<node> v1 = convex(a), v2 = convex(b);
	puts(check(v1, v2) ? "No" : "Yes");
}

int main() {
	while (scanf("%d%d", &n, &m) == 2 && n) {
		solve();
	}
	return 0;
}