[洛谷P5166]xtq的口令

题目大意：给出一张有向图，保证任何时候边都是从编号大的向编号小连。两个操作：

1. $1\;l\;r：$表示若编号在区间$[l,r]$内的点被染色了，问至少还需要染多少个点才可以使得整张图被染色。一个点会被染色的要求是：要么直接被染色，要么它所连向的点中至少一个被染色
2. $2\;l\;r\;x：$表示编号在区间$[l,r]$中的所有点都向$x$连一条边，保证$x<l$

题解：发现这张图是一个$DAG$，然后只要把所有出度为$0$的点染色就一定可以把所有点染色。

于是就记录每个点是否出度为$0$，询问是区间$[1,l)\cup(r,n]$中出度为$0$的点的个数，修改就把区间$[l,r]$中出度为$0$的点改为非$0$。可以用线段树维护

卡点：无

 

C++ Code：

#include <cstdio>
#include <cctype>
namespace std {
	struct istream {
#define M (1 << 24 | 3)
		char buf[M], *ch = buf - 1;
		inline istream() {
#ifndef ONLINE_JUDGE
			freopen("input.txt", "r", stdin);
#endif
			fread(buf, 1, M, stdin);
		}
		inline istream& operator >> (int &x) {
			while (isspace(*++ch));
			for (x = *ch & 15; isdigit(*++ch); ) x = x * 10 + (*ch & 15);
			return *this;
		}
#undef M
	} cin;
	struct ostream {
#define M (1 << 23 | 3)
		char buf[M], *ch = buf - 1;
		inline ostream& operator << (int x) {
			if (!x) {
				*++ch = '0';
				return *this;
			}
			static int S[20], *top; top = S;
			while (x) {
				*++top = x % 10 ^ 48;
				x /= 10;
			}
			for (; top != S; --top) *++ch = *top;
			return *this;
		}
		inline ostream& operator << (const char x) { *++ch = x; return *this; }
		inline ~ostream() {
#ifndef ONLINE_JUDGE
			freopen("output.txt", "w", stdout);
#endif
			fwrite(buf, 1, ch - buf + 1, stdout);
		}
#undef M
	} cout;
}

#define maxn 300010
int n, m;
bool notice[maxn];

namespace SgT {
	bool tg[maxn << 2];
	int V[maxn << 2];
	void build(int rt, int l, int r) {
		if (l == r) {
			V[rt] = !notice[l];
			return ;
		}
		int mid = l + r >> 1;
		build(rt << 1, l, mid), build(rt << 1 | 1, mid + 1, r);
		V[rt] = V[rt << 1] + V[rt << 1 | 1];
	}

	int L, R;
	inline void pushdown(int rt) {
		V[rt << 1] = V[rt << 1 | 1] = 0;
		tg[rt << 1] = tg[rt << 1 | 1] = true;
		tg[rt] = false;
	}
	void __modify(const int rt, const int l, const int r) {
		if (L <= l && R >= r) {
			V[rt] = 0;
			tg[rt] = true;
			return ;
		}
		if (tg[rt]) pushdown(rt);
		const int mid = l + r >> 1;
		if (L <= mid) __modify(rt << 1, l, mid);
		if (R > mid) __modify(rt << 1 | 1, mid + 1, r);
		V[rt] = V[rt << 1] + V[rt << 1 | 1];
	}
	void modify(const int __L, const int __R) {
		L = __L, R = __R;
		__modify(1, 1, n);
	}

	int res;
	void __query(const int rt, const int l, const int r) {
		if (L <= l && R >= r) {
			res += V[rt];
			return ;
		}
		if (tg[rt]) pushdown(rt);
		const int mid = l + r >> 1;
		if (L <= mid) __query(rt << 1, l, mid);
		if (R > mid) __query(rt << 1 | 1, mid + 1, r);
	}
	int query(const int __L, const int __R) {
		if (__L > __R) return 0;
		L = __L, R = __R, res = 0;
		__query(1, 1, n);
		return res;
	}
}

int main() {
	std::cin >> n >> m;
	for (int i = 1, k; i <= n; ++i) {
		std::cin >> k;
		for (int j = 0, x; j < k; ++j) std::cin >> x, notice[x] = true;
	}
	SgT::build(1, 1, n);
	while (m --> 0) {
		static int op, l, r, x;
		std::cin >> op >> l >> r;
		if (op == 1) std::cout << SgT::query(1, l - 1) + SgT::query(r + 1, n) << '\n';
		else std::cin >> x, SgT::modify(l, r);
	}
	return 0;
}