UOJ#191. 【集训队互测2016】Unknown

传送门
这个题目实际上可以建立出树，然后重链剖分维护一条链的凸包
然后离线询问排序斜率做到 (nlog^2n)，或者点分治+平衡树也行
但是这个题目卡空间，数组一不小心就爆了卡一卡也能过
考虑其它空间常数小并且又好写的做法
根据一般的二进制分组的方法，每次这个块满了就合并儿子的凸包
这样显然不对，只要又删又加就假了
我们换一种方法，每次这个块满了就合并线段树同一层前一个节点的儿子的凸包
这样每次都要花费 (len) 次添加操作才能换来一次 (2len) 的合并
此时的没有合并节点顶多 (2log) 个
查询就定位每个节点，在凸包上二分即可
时间 (nlog^2n) 空间 (nlogn)

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

namespace IO {
	const int maxn(1 << 21 | 1);

	char ibuf[maxn], *iS, *iT, c;
	int f;
	
	inline char Getc() {
		return iS == iT ? (iT = (iS = ibuf) + fread(ibuf, 1, maxn, stdin), (iS == iT ? EOF : *iS++)) : *iS++;
	}

	template <class Int> inline void In(Int &x) {
		for (c = Getc(), f = 1; c < '0' || c > '9'; c = Getc()) f = c == '-' ? -1 : 1;
		for (x = 0; c >= '0' && c <= '9'; c = Getc()) x = (x << 1) + (x << 3) + (c ^ 48);
		x *= f;
	}
}

using IO :: In;

const int maxn(1 << 20 | 1);
const int mod(998244353);
const ll inf(1e18);

struct Point {
	int x, y;

	inline Point(int _x = 0, int _y = 0) {
		x = _x, y = _y;
	}
	
	inline Point operator -(Point b) const {
		return Point(x - b.x, y - b.y);
	}

	inline ll operator *(Point b) const {
		return (ll)x * b.y - (ll)y * b.x;
	}
} cur;

int tp, m, mx, n, mxr;
bitset <maxn> done;
vector <Point> vc[maxn];

inline void Merge(int ls, int rs, int ff) {
	done[ff] = 1;
	int i, j, l1, l2, l3;
	vc[ff].clear(), l1 = vc[ls].size(), l2 = vc[rs].size();
	for (i = j = l3 = 0; i < l1 || j < l2; ) {
		if (j == l2 || (i < l1 && (vc[ls][i].x < vc[rs][j].x || (vc[ls][i].x == vc[rs][j].x && vc[ls][i].y < vc[rs][j].y)))) {
			while (l3 > 1 && (vc[ff][l3 - 1] - vc[ff][l3 - 2]) * (vc[ls][i] - vc[ff][l3 - 2]) >= 0) vc[ff].pop_back(), --l3;
			vc[ff].push_back(vc[ls][i]), ++i, ++l3;
		}
		else {
			while (l3 > 1 && (vc[ff][l3 - 1] - vc[ff][l3 - 2]) * (vc[rs][j] - vc[ff][l3 - 2]) >= 0) vc[ff].pop_back(), --l3;
			vc[ff].push_back(vc[rs][j]), ++j, ++l3;
		}
	}
}

inline ll Calc(int x) {
	if (!vc[x].size()) return -inf;
	int l, r, mid, ans;
	ans = vc[x].size() - 1, l = 0, r = vc[x].size() - 2;
	while (l <= r) {
		mid = (l + r) >> 1;
		if ((vc[x][mid + 1] - vc[x][mid]) * cur >= 0) ans = mid, r = mid - 1;
		else l = mid + 1;
	}
	return Point(cur.x, cur.y) * vc[x][ans];
}

void Insert(int x, int l, int r, int p) {
	int mid;
	if (l == r) {
		mxr = max(mxr, x);
		vc[x].push_back(cur), done[x] = 1;
		return;
	}
	mid = (l + r) >> 1;
	p <= mid ? Insert(x << 1, l, mid, p) : Insert(x << 1 | 1, mid + 1, r, p);
	if (x == (x & -x)) return;
	if (p == r && !done[x - 1]) Merge((x - 1) << 1, (x - 1) << 1 | 1, x - 1);
}

void Delete(int x, int l, int r, int p) {
	int mid;
	done[x] = 0;
	if (l == r) {
		mxr = max(mxr, x);
		vc[x].pop_back();
		return;
	}
	mid = (l + r) >> 1;
	p <= mid ? Delete(x << 1, l, mid, p) : Delete(x << 1 | 1, mid + 1, r, p);
}

ll Query(int x, int l, int r, int ql, int qr) {
	int mid;
	ll ret;
	if (ql <= l && qr >= r && done[x]) return Calc(x);
	if (l == r) return -inf;
	ret = -inf, mid = (l + r) >> 1;
	if (ql <= mid) ret = Query(x << 1, l, mid, ql, qr);
	if (qr > mid) ret = max(ret, Query(x << 1 | 1, mid + 1, r, ql, qr));
	return ret;
}

int main() {
	int i, op, x, y, l, r, ans;
	In(tp);
	while (In(m), m) {
		ans = n = 0, done.reset();
		for (i = 1; i <= mxr; ++i) vc[i].clear();
		for (mxr = 0, mx = 1; mx <= 300000; mx <<= 1);
		for (i = 1; i <= m; ++i) {
			In(op);
			if (op == 1) In(x), In(y), cur = Point(x, y), Insert(1, 1, mx, ++n);
			else if (op == 2) Delete(1, 1, mx, n), --n;
			else {
				In(l), In(r), In(x), In(y), cur = Point(x, y);
				ans ^= (Query(1, 1, mx, l, r) % mod + mod) % mod;
			}
		}
		printf("%d
", ans);
	}
    return 0;
}