【BZOJ5317】【JSOI2018】部落战争

【题目链接】

• 点击打开链接

【思路要点】

• 询问点\(c=(x,y)\)的答案为1当且仅当\(c\in \{a+(-b)|a\in A,b\in B\}\)。
• 求解两个点集凸包的闵可夫斯基和，然后判断询问点是否在求得的凸包中即可。
• 时间复杂度\(O(NLogN)\)。

【代码】

#include<bits/stdc++.h>
using namespace std;
const int MAXN = 3e5 + 5;
const double eps = 1e-11;
template <typename T> void read(T &x) {
	x = 0; int f = 1;
	char ch = getchar();
	for (; !isdigit(ch); ch = getchar()) if (ch == '-') f = -f;
	for (; isdigit(ch); ch = getchar()) x = x * 10 + ch - '0';
	x *= f;
}
struct point {int x, y; };
point operator + (point a, point b) {return (point) {a.x + b.x, a.y + b.y}; }
point operator - (point a, point b) {return (point) {a.x - b.x, a.y - b.y}; }
point operator * (point a, int b) {return (point) {a.x * b, a.y * b}; }
long long operator * (point a, point b) {return 1ll * a.x * b.y - 1ll * a.y * b.x; }
bool operator < (point a, point b) {
	if (a.y == b.y) return a.x < b.x;
	else return a.y < b.y;
}
long long dist(point a) {return 1ll * a.x * a.x + 1ll * a.y * a.y; }
int n, m, q;
point a[MAXN], b[MAXN], c[MAXN], fp;
bool cmp(point a, point b) {
	long long tmp = (a - fp) * (b - fp);
	if (tmp == 0) return dist(a - fp) < dist(b - fp);
	else return tmp > 0;
}
int main() {
	read(n), read(m), read(q);
	for (int i = 1; i <= n; i++)
		read(a[i].x), read(a[i].y);
	for (int i = 2; i <= n; i++)
		if (a[i] < a[1]) swap(a[i], a[1]);
	fp = a[1];
	sort(a + 2, a + n + 1, cmp);
	a[++n] = a[1];
	int top = 1;
	for (int i = 2; i <= n; i++) {
		while (top >= 2 && (a[top] - a[top - 1]) * (a[i] - a[top - 1]) <= 0) top--;
		a[++top] = a[i];
	}
	n = top;
	for (int i = 1; i <= m; i++) {
		read(b[i].x), read(b[i].y);
		b[i] = b[i] * (-1);
	}
	for (int i = 2; i <= m; i++)
		if (b[i] < b[1]) swap(b[i], b[1]);
	fp = b[1];
	sort(b + 2, b + m + 1, cmp);
	b[++m] = b[1]; top = 1;
	for (int i = 2; i <= m; i++) {
		while (top >= 2 && (b[top] - b[top - 1]) * (b[i] - b[top - 1]) <= 0) top--;
		b[++top] = b[i];
	}
	m = top;
	int pn = 1, pm = 1;
	c[1] = a[1] + b[1];
	for (int i = 2; i <= n + m - 1; i++) {
		if (pn == n) {
			point tmp = b[pm + 1] - b[pm]; pm++;
			c[i] = c[i - 1] + tmp;
		} else if (pm == m) {
			point tmp = a[pn + 1] - a[pn]; pn++;
			c[i] = c[i - 1] + tmp;
		} else {
			point tmp = a[pn + 1] - a[pn], tnp = b[pm + 1] - b[pm];
			if (tmp * tnp >= 0) c[i] = c[i - 1] + tmp, pn++;
			else c[i] = c[i - 1] + tnp, pm++;
		}
	}
	n = n + m - 1;
	top = 1;
	for (int i = 2; i <= n; i++) {
		while (top >= 2 && (c[top] - c[top - 1]) * (c[i] - c[top - 1]) <= 0) top--;
		c[++top] = c[i];
	}
	n = top;
	for (int i = 2; i <= n - 1; i++)
		if (c[i].y >= c[i - 1].y && c[i].y >= c[i + 1].y) {
			m = i;
			break;
		}
	for (int i = 1; i <= q; i++) {
		point pos; read(pos.x), read(pos.y);
		if (pos.y < c[1].y || pos.y > c[m].y) {
			printf("0\n");
			continue;
		}
		if (pos.y == c[1].y) {
			int vl = c[1].x, vr = c[1].x;
			if (pos.y == c[2].y) vr = c[2].x;
			printf("%d\n", pos.x >= vl && pos.x <= vr);
			continue;
		}
		if (pos.y == c[m].y) {
			int vl = c[m].x, vr = c[m].x;
			if (pos.y == c[m + 1].y) vl = c[m + 1].x;
			printf("%d\n", pos.x >= vl && pos.x <= vr);
			continue;
		}
		double vl, vr;
		int l = 1, r = m;
		while (l < r) {
			int mid = (l + r + 1) / 2;
			if (c[mid].y <= pos.y) l = mid;
			else r = mid - 1;
		}
		vr = c[l].x + (double) (c[l + 1].x - c[l].x) / (c[l + 1].y - c[l].y) * (pos.y - c[l].y);
		l = m, r = n;
		while (l < r) {
			int mid = (l + r + 1) / 2;
			if (c[mid].y >= pos.y) l = mid;
			else r = mid - 1;
		}
		vl = c[l].x + (double) (c[l + 1].x - c[l].x) / (c[l + 1].y - c[l].y) * (pos.y - c[l].y);
		printf("%d\n", pos.x >= vl - eps && pos.x <= vr + eps);
	}
	return 0;
}