C++11(clang++ 3.9) 解法, 执行用时: 282ms, 内存消耗: 844K, 提交时间: 2020-07-01 18:11:47

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

inline int read() {
	char c = getchar(); int x = 0, f = 1;
	while (c < '0' || c > '9') { if (c == '-') f = -1; c = getchar(); }
	while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
	return x * f;
}

struct Node_t {
	int l, r; mutable int v;
	bool operator < (const Node_t &o) const { return l < o.l; } 
};

set<Node_t> odt;

int n, m, pl, pr;

inline set<Node_t>::iterator Split(int x) {
	if (x > n) return odt.end();
	auto it = --odt.upper_bound({ x, 0, 0 });
	if (it->l == x) return it;
	int l = it->l, r = it->r, v = it->v;
	odt.erase(it); odt.insert({ l, x - 1, v });
	return odt.insert({ x, r, v }).first;
}
inline void Assign(int l, int r, int v) {
	auto itr = Split(r + 1), itl = Split(l);
	odt.erase(itl, itr); odt.insert({ l, r, v });
}
inline int Count(int l, int r) {
	auto itr = Split(r + 1), itl = Split(l); int ret = 0;
	for (; itl != itr; itl++) if (itl->v == 1) ret += itl->r - itl->l + 1;
	return ret;
}
inline void Modify(int l, int r, int p) {
	auto itr = Split(r + 1), itl = Split(l); int ret = 0;
	if (!p) return;
	for (; itl != itr; itl++)
		if (!(itl->v)) {
			int now = itl->r - itl->l + 1;
			if (now <= p) { p -= now; itl->v = 1; }
			else { pl = itl->l; pr = pl + p - 1; p = 0; }
			if (!p) return;
		}
}
inline int Query(int l, int r) {
	auto itr = Split(r + 1), itl = Split(l); int Mx = 0, lst = 0;
	for (; itl != itr; itl++)
		if (!(itl->v)) {
			lst += itl->r - itl->l + 1; Mx = max(Mx, lst);
		} else lst = 0;
	return Mx;
}

int main() {
	n = read(); m = read();
	odt.insert({ 1, n, 1 });
	while (m--) {
		int opt = read(), l = read(), r = read();
		if (!opt) Assign(l, r, 0);
		else if (opt == 1) {
			int x = read(), y = read(), p = Count(l, r);
			pl = pr = -1;
			Assign(l, r, 0); Modify(x, y, p);
			if (pl != -1 && pr != -1) Assign(pl, pr, 1);
		} else printf("%d\n", Query(l, r));
	}
	return 0;
}