【BZOJ4942】【UOJ314】【NOI2017】整数

【题目链接】

【思路要点】

  • 用线段树模拟整数的加减,通过维护区间信息和区间赋值来实现快速进/退位。
  • 具体而言,在每个节点维护该区间是否全为0/1,以及区间赋值懒标记。
  • 进行加/减法时现在对应位上进行操作,若涉及进/退位,则在线段树上二分出最近的更高位的0/1,对其进行操作,并将操作位置和该位置之间的区间赋值为1/0。
  • 进行压位,进制变为\(2^{30}\)。时间复杂度\(O(NLogN)\)。

【代码】

#include<bits/stdc++.h>
using namespace std;
const int MAXN = 1500005;
const int MAXP = 2500005;
const int s = (1 << 30) - 1;
template <typename T> void chkmax(T &x, T y) {x = max(x, y); }
template <typename T> void chkmin(T &x, T y) {x = min(x, y); } 
template <typename T> void read(T &x) {
	x = 0; int f = 1;
	char c = getchar();
	for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
	for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
	x *= f;
}
template <typename T> void write(T x) {
	if (x < 0) x = -x, putchar('-');
	if (x > 9) write(x / 10);
	putchar(x % 10 + '0');
}
template <typename T> void writeln(T x) {
	write(x);
	puts("");
}
struct SegmentTree {
	struct Node {
		int lc, rc;
		bool tag, tagtype;
		bool all, alltype;
	} a[MAXP];
	int root, size, n, val[MAXN];
	void pushdown(int root, int l, int r) {
		if (!a[root].tag) return;
		if (l == r) {
			if (a[root].tagtype) val[l] = s;
			else val[l] = 0;
		} else {
			int tmp = a[root].lc;
			a[tmp].tag = true;
			a[tmp].tagtype = a[root].tagtype;
			a[tmp].all = true;
			a[tmp].alltype = a[root].tagtype;
			tmp = a[root].rc;
			a[tmp].tag = true;
			a[tmp].tagtype = a[root].tagtype;
			a[tmp].all = true;
			a[tmp].alltype = a[root].tagtype;
		}
		a[root].tag = false;
	}
	void update(int root, int l, int r) {
		if (l == r) {
			if (val[l] == s) {
				a[root].all = true;
				a[root].alltype = true;
			} else if (val[l] == 0) {
				a[root].all = true;
				a[root].alltype = false;
			} else a[root].all = false;
		} else {
			if (a[a[root].lc].all && a[a[root].rc].all && a[a[root].lc].alltype == a[a[root].rc].alltype) {
				a[root].all = true;
				a[root].alltype = a[a[root].lc].alltype;
			} else a[root].all = false;
		}
	}
	void build(int &root, int l, int r) {
		root = ++size;
		if (l == r) {
			update(root, l, r);
			return;
		}
		int mid = (l + r) / 2;
		build(a[root].lc, l, mid);
		build(a[root].rc, mid + 1, r);
		update(root, l, r);
	}
	void init(int x) {
		n = x;
		build(root, 0, n);
	}
	void access(int root, int l, int r, int pos) {
		pushdown(root, l, r);
		if (l == r) {
			update(root, l, r);
			return;
		}
		int mid = (l + r) / 2;
		if (mid >= pos) access(a[root].lc, l, mid, pos);
		else access(a[root].rc, mid + 1, r, pos);
		update(root, l, r);
	}
	int query(int pos, int bit) {
		access(root, 0, n, pos);
		return (val[pos] & (1 << bit)) != 0;
	}
	bool add(int root, int l, int r, int pos) {
		pushdown(root, l, r);
		int mid = (l + r) / 2;
		if (l > pos) {
			if (a[root].all && a[root].alltype) {
				a[root].alltype = false;
				a[root].tag = true;
				a[root].tagtype = false;
				return false;
			}
			if (l == r) {
				val[l]++;
				update(root, l, r);
				return true;
			}
			if (add(a[root].lc, l, mid, pos)) {
				update(root, l, r);
				return true;
			}
			bool ans = add(a[root].rc, mid + 1, r, pos);
			update(root, l, r);
			return ans;
		}
		if (mid > pos && add(a[root].lc, l, mid, pos)) {
			update(root, l, r);
			return true;
		}
		bool ans = add(a[root].rc, mid + 1, r, pos);
		update(root, l, r);
		return ans;
	}
	void add(int pos, int d) {
		access(root, 0, n, pos);
		val[pos] += d;
		if (val[pos] <= s) {
			access(root, 0, n, pos);
			return;
		}
		val[pos] &= s;
		access(root, 0, n, pos);
		add(root, 0, n, pos);
	}
	bool dec(int root, int l, int r, int pos) {
		pushdown(root, l, r);
		int mid = (l + r) / 2;
		if (l > pos) {
			if (a[root].all && !a[root].alltype) {
				a[root].alltype = true;
				a[root].tag = true;
				a[root].tagtype = true;
				return false;
			}
			if (l == r) {
				val[l]--;
				update(root, l, r);
				return true;
			}
			if (dec(a[root].lc, l, mid, pos)) {
				update(root, l, r);
				return true;
			}
			bool ans = dec(a[root].rc, mid + 1, r, pos);
			update(root, l, r);
			return ans;
		}
		if (mid > pos && dec(a[root].lc, l, mid, pos)) {
			update(root, l, r);
			return true;
		}
		bool ans = dec(a[root].rc, mid + 1, r, pos);
		update(root, l, r);
		return ans;
	}
	void dec(int pos, int d) {
		access(root, 0, n, pos);
		val[pos] -= d;
		if (val[pos] >= 0) {
			access(root, 0, n, pos);
			return;
		}
		val[pos] += 1 << 30;
		access(root, 0, n, pos);
		dec(root, 0, n, pos);
	}
} ST;
int n, t1, t2, t3;
int main() {
	read(n), read(t1), read(t2), read(t3);
	ST.init(n + 1840);
	for (int i = 1; i <= n; i++) {
		int opt; read(opt);
		if (opt == 1) {
			long long x; int y;
			read(x), read(y);
			int pos = y / 30, bit = y % 30;
			if (x >= 0) {
				x <<= bit;
				ST.add(pos, x & s);
				ST.add(pos + 1, x >> 30);
			} else {
				x = -x;
				x <<= bit;
				ST.dec(pos, x & s);
				ST.dec(pos + 1, x >> 30);
			}
		} else {
			int x; read(x);
			writeln(ST.query(x / 30, x % 30));
		}
	}
	return 0;
}

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转载自blog.csdn.net/qq_39972971/article/details/80513830