[洛谷P2824][HEOI2016/TJOI2016]排序

题目大意：一个全排列，两种操作：

1. $0\;l\;r：$把$[l,r]$升序排序
2. $1\;l\;r：$把$[l,r]$降序排序

最后询问第$k$位是什么

题解：二分答案，把比这个数大的赋成$1$，否则为$0$，线段树区间和和区间赋$01$，最后判断第$k$位是$0$是$1$，若为$1$则还可以变大，否则变小

卡点：修改后没有$update$

C++ Code：

#include <cstdio>
#include <cctype>
namespace __R {
	int x, ch;
	inline int read() {
		ch = getchar();
		while (isspace(ch)) ch = getchar();
		for (x = ch & 15, ch = getchar(); isdigit(ch); ch = getchar()) x = x * 10 + (ch & 15);
		return x;
	}
}
using __R::read;

#define maxn 100010

int n, m, q;
int s[maxn], op[maxn], L[maxn], R[maxn];
namespace SgT {
	int V[maxn << 2], tg[maxn << 2];
	int num, L, R;

	void build(int rt, int l, int r) {
		tg[rt] = -1;
		if (l == r) {
			V[rt] = s[l] > num;
			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];
	}
	void build(int __num) {
		num = __num;
		build(1, 1, n);
	}

	inline void pushdown(int rt, int len) {
		int &__tg = tg[rt];
		V[rt << 1] = (len + 1 >> 1) * __tg;
		tg[rt << 1] = __tg;
		V[rt << 1 | 1] = (len >> 1) * __tg;
		tg[rt << 1 | 1] = __tg;
		__tg = -1;
	}
	int __query(int rt, int l, int r) {
		if (L <= l && R >= r) return V[rt];
		int mid = l + r >> 1, ans = 0;
		if (~tg[rt]) pushdown(rt, r - l + 1);
		if (L <= mid) ans = __query(rt << 1, l, mid);
		if (R > mid) ans += __query(rt << 1 | 1, mid + 1, r);
		return ans;
	}
	int query(int __L, int __R) {
		L = __L, R = __R;
		return __query(1, 1, n);
	}

	void __modify(int rt, int l, int r) {
		if (L <= l && R >= r) {
			V[rt] = num * (r - l + 1);
			tg[rt] = num;
			return ;
		}
		int mid = l + r >> 1;
		if (~tg[rt]) pushdown(rt, r - l + 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(int __L, int __R, int __num) {
		L = __L, R = __R, num = __num;
		__modify(1, 1, n);
	}
}

bool check(int mid) {
	SgT::build(mid);
	for (int i = 1; i <= m; i++) {
		int _1 = SgT::query(L[i], R[i]), _0 = R[i] - L[i] + 1 - _1;
		if (op[i]) {
			SgT::modify(L[i], L[i] + _1 - 1, 1);
			SgT::modify(L[i] + _1, R[i], 0);
		} else {
			SgT::modify(L[i], L[i] + _0 - 1, 0);
			SgT::modify(L[i] + _0, R[i], 1);
		}
	}
	return SgT::query(q, q);
}

int main() {
	n = read(), m = read();
	for (int i = 1; i <= n; i++) s[i] = read();
	for (int i = 1; i <= m; i++) {
		op[i] = read(), L[i] = read(), R[i] = read();
	}
	q = read();
	int l = 1, r = n, ans = 1;
	while (l <= r) {
		int mid = l + r >> 1;
		if (check(mid)) l = mid + 1;
		else {
			r = mid - 1;
			ans = mid;
		}
	}
	printf("%d\n", ans);
	return 0;
}