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AtCoder Beginner Contest 251 G Intersection of Polygons

发布时间 2023-06-16 10:41:16作者: zltzlt

洛谷传送门

AtCoder 传送门

经典结论，一个点 \(P(x, y)\) 在一个凸多边形内部 \(S = \{(x_i, y_i)\}\) 的充要条件是 \(\forall i \in [1, n], (x_{i + 1} - x_i, y_{i + 1} - y_i) \times (x - x_i, y - y_i) \ge 0\)，其中 \(S\) 的点按照逆时针排列。

然后我们运用叉积的一个性质 \((x_{i + 1} - x_i, y_{i + 1} - y_i) \times (x - x_i, y - y_i) = (x_{i + 1} - x_i, y_{i + 1} - y_i) \times (x, y) - (x_{i + 1} - x_i, y_{i + 1} - y_i) \times (x_i, y_i)\) 就做完了，预处理后项最大值即可。

code 
// Problem: G - Intersection of Polygons
// Contest: AtCoder - Panasonic Programming Contest 2022(AtCoder Beginner Contest 251)
// URL: https://atcoder.jp/contests/abc251/tasks/abc251_g
// Memory Limit: 1024 MB
// Time Limit: 2000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

#include <bits/stdc++.h>
#define pb emplace_back
#define fst first
#define scd second
#define mems(a, x) memset((a), (x), sizeof(a))

using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef double db;
typedef long double ldb;
typedef pair<ll, ll> pii;

const int maxn = 200100;

ll n, m, q, a[maxn], b[maxn], c[maxn], d[maxn], e[maxn], f[maxn], g[maxn];

inline ll calc(ll a, ll b, ll c, ll d) {
	return a * d - b * c;
}

void solve() {
	scanf("%lld", &n);
	for (int i = 1; i <= n; ++i) {
		scanf("%lld%lld", &a[i], &b[i]);
	}
	scanf("%lld", &m);
	for (int i = 1; i <= m; ++i) {
		scanf("%lld%lld", &c[i], &d[i]);
	}
	a[n + 1] = a[1];
	b[n + 1] = b[1];
	scanf("%lld", &q);
	for (int i = 1; i <= n; ++i) {
		e[i] = a[i + 1] - a[i];
		f[i] = b[i + 1] - b[i];
		g[i] = -9e18;
		for (int j = 1; j <= m; ++j) {
			g[i] = max(g[i], calc(e[i], f[i], a[i] + c[j], b[i] + d[j]));
		}
	}
	while (q--) {
		ll x, y;
		scanf("%lld%lld", &x, &y);
		bool flag = 1;
		for (int i = 1; i <= n; ++i) {
			if (calc(e[i], f[i], x, y) < g[i]) {
				flag = 0;
				break;
			}
		}
		puts(flag ? "Yes" : "No");
	}
}

int main() {
	int T = 1;
	// scanf("%d", &T);
	while (T--) {
		solve();
	}
	return 0;
}