Solution -「HEOI/TJOI 2016」「洛谷 P2824」排序

$\mathcal{Description}$

  Link.

  给定排列 $\{p_n\}$ 和 $m$ 次局部排序操作，求操作完成后第 $q$ 位的值。

  $n,m\le10^5$。

$\mathcal{Solution}$

  跟这道的核心套路（？）差不多。

  若序列是 $01$ 序列，局部排序就相当于把 $1$ 扔到一端，把 $0$ 扔到另一端，只需要知道区间 $1$ 的个数就好。

  二分答案 $mid$，将排列中不小于 $mid$ 的值设为 $1$，其余设为 $0$，暴力建新的线段树维护区间和，然后暴力处理每次排序操作，最后求到此时 $q$ 位置的值（$1$ 或 $0$）。注意到这个值的意义——$q$ 位置的值大于等于 / 小于 $mid$，借此调整二分区间即可。

  复杂度 $\mathcal O(m\log^2n+n\log n)$。

$\mathcal{Code}$

#include <cstdio>

const int MAXN = 1e5;
int n, m, a[MAXN + 5]; 

struct Event { int op, l, r; } evt[MAXN + 5];

inline int rint () {
	int x = 0; char s = getchar ();
	for ( ; s < '0' || '9' < s; s = getchar () );
	for ( ; '0' <= s && s <= '9'; s = getchar () ) x = x * 10 + ( s ^ '0' );
	return x;
}

struct SegmentTree {
	int one[MAXN << 2], tag[MAXN << 2];

	inline void pushup ( const int rt ) { one[rt] = one[rt << 1] + one[rt << 1 | 1]; }

	inline void pushdn ( const int rt, const int len ) {
		if ( ! ~ tag[rt] ) return ;
		one[rt << 1] = tag[rt] * ( len + 1 >> 1 );
		one[rt << 1 | 1] = tag[rt] * ( len >> 1 );
		tag[rt << 1] = tag[rt << 1 | 1] = tag[rt];
		tag[rt] = -1;
	}

	inline void build ( const int rt, const int l, const int r, const int thrval ) {
		tag[rt] = -1;
		if ( l == r ) return void ( one[rt] = a[l] >= thrval );
		int mid = l + r >> 1;
		build ( rt << 1, l, mid, thrval ), build ( rt << 1 | 1, mid + 1, r, thrval );
		pushup ( rt );
	}

	inline void assign ( const int rt, const int l, const int r, const int al, const int ar, const int v ) {
		if ( al > ar ) return ;
		if ( al <= l && r <= ar ) return void ( one[rt] = ( tag[rt] = v ) * ( r - l + 1 ) );
		int mid = l + r >> 1; pushdn ( rt, r - l + 1 );
		if ( al <= mid ) assign ( rt << 1, l, mid, al, ar, v );
		if ( mid < ar ) assign ( rt << 1 | 1, mid + 1, r, al, ar, v );
		pushup ( rt );
	}

	inline int query ( const int rt, const int l, const int r, const int ql, const int qr ) {
		if ( ql <= l && r <= qr ) return one[rt];
		int mid = l + r >> 1, ret = 0; pushdn ( rt, r - l + 1 );
		if ( ql <= mid ) ret += query ( rt << 1, l, mid, ql, qr );
		if ( mid < qr ) ret += query ( rt << 1 | 1, mid + 1, r, ql, qr );
		return ret;
	}
} st;

int main () {
	n = rint (), m = rint ();
	for ( int i = 1; i <= n; ++ i ) a[i] = rint ();
	for ( int i = 1; i <= m; ++ i ) {
		evt[i].op = rint (), evt[i].l = rint (), evt[i].r = rint ();
	}
	int l = 1, r = n, q = rint ();
	while ( l < r ) {
		int mid = l + r + 1 >> 1;
		st.build ( 1, 1, n, mid );
		for ( int i = 1; i <= m; ++ i ) {
			int el = evt[i].l, er = evt[i].r, t = st.query ( 1, 1, n, el, er );
			if ( ! evt[i].op ) {
				st.assign ( 1, 1, n, el, er - t, 0 );
				st.assign ( 1, 1, n, er - t + 1, er, 1 );
			} else {
				st.assign ( 1, 1, n, el, el + t - 1, 1 );
				st.assign ( 1, 1, n, el + t, er, 0 );
			}
		}
		if ( st.query ( 1, 1, n, q, q ) ) l = mid;
		else r = mid - 1;
	}
	printf ( "%d\n", l );
	return 0;
}