「JSOI2018」战争

解题思路

我们需要每次求给一个凸包加上一个向量后是否与另外一个凸包相交，也就是说是否存在

这里 \(A, B\) 表示凸包内部的点集，可以转化一步变成

那相当于对 \(A,(-B)\) 作闵可夫斯基和，判断 \(w\) 是否在新的凸包内部，把新的凸包划分成三角区域，让 \(w\) 和原点做一条向量，二分一下在哪个区域然后判断一下在区域内部还是外部就可以了，复杂度 \(\mathcal O(n \log n)\) 。

code

/*program by mangoyang*/

#include<bits/stdc++.h>

#define inf (0x7f7f7f7f)

#define Max(a, b) ((a) > (b) ? (a) : (b))

#define Min(a, b) ((a) < (b) ? (a) : (b))

typedef long long ll;

using namespace std;

template <class T>

inline void read(T &x){

    int ch = 0, f = 0; x = 0;

    for(; !isdigit(ch); ch = getchar()) if(ch == '-') f = 1;

    for(; isdigit(ch); ch = getchar()) x = x * 10 + ch - 48;

    if(f) x = -x;

}

const int N = 1000005;

struct P{

	ll x, y;

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

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

	friend ll operator * (P a, P b){ return a.x * b.y - b.x * a.y; }

	inline ll dis(){ return x * x + y * y; }

}A[N], B[N], C[N], s1[N], s2[N], st[N], O;

int n, m, q;

inline bool cmp1 (P A, P B){

	return A.y != B.y ? A.y < B.y : A.x < B.x;

}

inline bool cmp2 (P A, P B){

	//叉积一样按照离原点距离排，防止较远的点被近的点日掉

	ll res = (A - O) * (B - O);

	return res ? res > 0 : (A - O).dis() < (B - O).dis();

}

inline int convex(P *A, int len){

	//求点集 A 的凸包并返回凸包大小

	sort(A + 1, A + len + 1, cmp1); O = A[1];

	sort(A + 2, A + len + 1, cmp2);

	int top = 1; st[top] = A[1];

	for(int i = 2; i <= len; i++){

		while(top > 1 && (st[top] - st[top-1]) * (A[i] - st[top-1]) <= 0) top--;

		st[++top] = A[i];

	}

	for(int i = 1; i <= top; i++) A[i] = st[i];

	return top;

}

inline int inconvex(P x, P *A, int len){

	//判断点 x 是否在大小为len的凸包 A 里，二分找到向量所在的三角区域

	O = A[1];

	if((x - O) * (A[2] - O) > 0 || (x - O) * (A[len] - O) < 0) return 0;

	int pos = lower_bound(A + 2, A + len + 1, x, cmp2) - A - 1;

	return (x - A[pos]) * (A[pos%len+1] - A[pos]) <= 0;

}

inline int Minkowski(P *A, P *B, P *C, int n, int m){

	//将大小为 n, m 的凸包 A, B 的闵可夫斯基和存在 C 中，并返回凸包大小

	int tot1 = 0, tot2 = 0;

	for(int i = 1; i < n; i++) s1[++tot1] = A[i+1] - A[i];

	s1[++tot1] = A[1] - A[n];

	for(int i = 1; i < m; i++) s2[++tot2] = B[i+1] - B[i];

	s2[++tot2] = B[1] - B[m];

	int p1 = 1, p2 = 1, tot = 1; C[tot] = A[1] + B[1];

	for(; p1 <= n && p2 <= m; tot++)

		C[tot+1] = C[tot] + (s1[p1] * s2[p2] >= 0 ? s1[p1++] : s2[p2++]);

	for(; p1 <= n; p1++, tot++) C[tot+1] = C[tot] + s1[p1];

	for(; p2 <= m; p2++, tot++) C[tot+1] = C[tot] + s2[p2];

	return tot = convex(C, tot);

}

int main(){

	read(n), read(m), read(q);

	for(int i = 1; i <= n; i++) read(A[i].x), read(A[i].y);

	n = convex(A, n);

	for(int i = 1; i <= m; i++)

		read(B[i].x), read(B[i].y), B[i].x = -B[i].x, B[i].y = -B[i].y;

	m = convex(B, m);

	int len = Minkowski(A, B, C, n, m);

	while(q--){

		ll x, y; read(x), read(y);

		printf("%d\n", inconvex((P){x, y}, C, len));

	}

	return 0;

}