Solution -「Ynoi 2018」「洛谷 P4117」五彩斑斓的世界

$\mathcal{Description}$

  Link.

  给定序列 $\{a_n\}$，处理 $m$ 次操作：

1. 给定 $l,r,x$，把 $[l,r]$ 内所有 $>x$ 的数减去 $x$；
2. 给定 $l,r,x$，查询 $[l,r]$ 内 $x$ 的出现次数。

  $n\le10^6$，$m\le5\times10^5$，$0\le a_i,x\le10^5$。

$\mathcal{Solution}$

  巧妙的分块题。

  分块，设块长为 $B$，对于每一块，利用值域 $V$ 不大，直接使用桶的方式维护信息。把桶看做一条线段 $[v_l,v_r]$，每次修改相当于把线段的后端截下一段平移到前端，可用启发式合并 + 并查集确保复杂度。最终复杂度为 $\mathcal O(qB+\frac{nV}{B}+\frac{qn}{B})$，大概取 $B=1.5\times10^3$ 就好。注意卡空间，对每个块离线处理。

$\mathcal{Code}$

/*~Rainybunny~*/

#include <bits/stdc++.h>

#define rep( i, l, r ) for ( int i = l, rep##i = r; i <= rep##i; ++i )
#define per( i, r, l ) for ( int i = r, per##i = l; i >= per##i; --i )

inline char fgc() {
	static char buf[1 << 17], *p = buf, *q = buf;
	return p == q && ( q = buf + fread( p = buf, 1, 1 << 17, stdin ), p == q )
	  ? EOF : *p++;
}

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

inline void wint( const int x ) {
	if ( 9 < x ) wint( x / 10 );
	putchar( x % 10 ^ '0' );
}

inline int imin( const int u, const int v ) { return u < v ? u : v; }
inline int imax( const int u, const int v ) { return u < v ? v : u; }

const int MAXN = 1e6, MAXM = 6e5, MAXV = 1e5, MAXS = 1500;
int n, m, a[MAXN + 5], ans[MAXM + 5];
struct Event { int op, l, r, x; } evt[MAXM + 5];
int fa[MAXN + 5], siz[MAXV + 5], rt[MAXV + 5], ref[MAXN + 5];

inline int find( const int x ) {
	return x == fa[x] ? x : fa[x] = find( fa[x] );
}

inline void unite( int x, int y ) {
	assert( rt[x] );
	if ( rt[y] ) fa[rt[x]] = rt[y];
	else ref[rt[y] = rt[x]] = y;
	siz[y] += siz[x], siz[x] = rt[x] = 0;
}

inline void solve( const int bl, const int br ) {
	int vl = 0, vr = 0;
	
	rep ( i, bl, br ) vr = a[i] < vr ? vr : a[i];
	memset( rt, 0, sizeof rt ), memset( siz, 0, sizeof siz );
	rep ( i, bl, br ) {
		if ( rt[a[i]] ) fa[i] = rt[a[i]];
		else rt[a[i]] = fa[i] = i, ref[i] = a[i];
		++siz[a[i]];
	}
	
	rep ( i, 1, m ) {
		int x = evt[i].x, l = evt[i].l, r = evt[i].r;
		if ( r < bl || l > br ) continue;
		if ( evt[i].op == 1 ) { // modify.
			if ( vr - vl <= x ) continue;
			if ( l <= bl && br <= r ) { // whole block update.
				if ( x << 1 > vr - vl ) { // right to left.
					per ( j, vr, vl + x + 1 ) if ( rt[j] ) unite( j, j - x );
					vr = vl + x;
				} else { // left to right.
					rep ( j, vl + 1, vl + x ) if ( rt[j] ) unite( j, j + x );
					vl += x;
				}
			} else { // partly update. curA = find(orgA)-vl.
				rep ( j, bl, br ) {
					a[j] = ref[find( j )];
					siz[a[j]] = rt[a[j]] = 0;
					a[j] -= vl;
				}
				
				vl = vr = 0;
				rep ( j, bl, br ) {
					l <= j && j <= r && a[j] > x && ( a[j] -= x );
					vr = vr < a[j] ? a[j] : vr;
					if ( rt[a[j]] ) fa[j] = rt[a[j]];
					else rt[a[j]] = fa[j] = j, ref[j] = a[j];
					++siz[a[j]];
				}
			}
		} else if ( vl + x <= vr ) { // meaningful query.
			if ( l <= bl && br <= r ) { // whole block query.
				ans[i] += siz[x + vl];
			} else { // partly query.
				rep ( j, imax( l, bl ), imin( r, br ) ) {
					ans[i] += ref[find( j )] - vl == x;
				}
			}
		}
	}
}

int main() {
    n = rint(), m = rint();
	rep ( i, 1, n ) a[i] = rint();
	rep ( i, 1, m ) {
		evt[i].op = rint();
		evt[i].l = rint(), evt[i].r = rint(), evt[i].x = rint();
	}
	
	for ( int l = 1, r; l <= n; l = r + 1 ) {
		r = l + MAXS - 1; if ( r > n ) r = n;
		solve( l, r );
	}
	
	rep ( i, 1, m ) if ( evt[i].op == 2 ) {
		wint( ans[i] ), putchar( '\n' );
	}
	return 0;
}