CF1093E Intersection of Permutations [分块 +bitset]

大家好， 我非常喜欢暴力数据结构， 于是就用分块A了此题

分块题，考虑前缀和 \(b_i\) 表示 bitset 即 \(0\) ~ \(i
\) 出现过的数字，然后考虑直接暴力复制块然后前缀和，修改也很简单，直接删除完了 swap 一下，再弄回去

时间复杂度 \(O(q \sqrt n + q \frac{n}{w})\)

// by Isaunoya
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#include <cstdio>
#include <bitset>

using namespace std;

#define rep(i, x, y) for (register int i = (x); i <= (y); ++i)
#define Rep(i, x, y) for (register int i = (x); i >= (y); --i)
//#define int long long

const int _ = 1 << 21;
struct I {
	char fin[_], *p1 = fin, *p2 = fin;
	inline char gc() {
		return (p1 == p2) && (p2 = (p1 = fin) + fread(fin, 1, _, stdin), p1 == p2) ? EOF : *p1++;
	}
	inline I& operator>>(int& x) {
		bool sign = 1;
		char c = 0;
		while (c < 48) ((c = gc()) == 45) && (sign = 0);
		x = (c & 15);
		while ((c = gc()) > 47) x = (x << 1) + (x << 3) + (c & 15);
		x = sign ? x : -x;
		return *this;
	}
	inline I& operator>>(double& x) {
		bool sign = 1;
		char c = 0;
		while (c < 48) ((c = gc()) == 45) && (sign = 0);
		x = (c - 48);
		while ((c = gc()) > 47) x = x * 10 + (c - 48);
		if (c == '.') {
			double d = 1.0;
			while ((c = gc()) > 47) d = d * 0.1, x = x + (d * (c - 48));
		}
		x = sign ? x : -x;
		return *this;
	}
	inline I& operator>>(char& x) {
		do
			x = gc();
		while (isspace(x));
		return *this;
	}
	inline I& operator>>(string& s) {
		s = "";
		char c = gc();
		while (isspace(c)) c = gc();
		while (!isspace(c) && c != EOF) s += c, c = gc();
		return *this;
	}
} in;
struct O {
	char st[100], fout[_];
	signed stk = 0, top = 0;
	inline void flush() {
		fwrite(fout, 1, top, stdout), fflush(stdout), top = 0;
	}
	inline O& operator<<(int x) {
		if (top > (1 << 20)) flush();
		if (x < 0) fout[top++] = 45, x = -x;
		do
			st[++stk] = x % 10 ^ 48, x /= 10;
		while (x);
		while (stk) fout[top++] = st[stk--];
		return *this;
	}
	inline O& operator<<(char x) {
		fout[top++] = x;
		return *this;
	}
	inline O& operator<<(string s) {
		if (top > (1 << 20)) flush();
		for (char x : s) fout[top++] = x;
		return *this;
	}
} out;
#define pb emplace_back
#define fir first
#define sec second

template < class T > inline void cmax(T & x , const T & y) {
	(x < y) && (x = y) ;
}
template < class T > inline void cmin(T & x , const T & y) {
	(x > y) && (x = y) ;
}

int n , m ;
const int maxn = 2e5 + 200 ;
template < int bl >
struct BLOCK {
	bitset < maxn > b[maxn / bl] ;
	bitset < maxn > temp ;
	int per[maxn] ;
	inline void ins(int x , int v) {
		per[x] = v ;
		for(register int i = x ; i <= n ; i += bl)
			b[i / bl].set(v) ;
	}
	inline void del(int x , int v)  {
		for(register int i = x ; i <= n ; i += bl)
			b[i / bl].reset(v) ;
	}
	inline bitset < maxn > qry(int x) {
		temp = x / bl ? b[x / bl - 1] : 0 ;
		rep(i , max(x / bl * bl , 1) , x)
			temp.set(per[i]) ;
		return temp ;
	}
	inline bitset < maxn > qry(int l , int r) {
		return qry(r) ^ qry(l - 1) ;
	}
};
int a[maxn] , b[maxn] ;
BLOCK < 256 > A ;
BLOCK < 512 > B ;
signed main() {
#ifdef _WIN64
	freopen("testdata.in" , "r" , stdin) ;
#endif
	in >> n >> m ;
	rep(i , 1 , n) in >> a[i] , A.ins(i , a[i]) ;
	rep(i , 1 , n) in >> b[i] , B.ins(i , b[i]) ;
	while(m --) {
		int opt ;
		in >> opt ;
		if(opt == 1) {
			int l , r , a , b ;
			in >> l >> r >> a >> b ;
			int ans = (A.qry(l , r) & B.qry(a , b)).count() ;
			out << ans << '\n' ;
		}
		else {
			int x , y ;
			in >> x >> y ;
			B.del(x , b[x]) ; B.del(y , b[y]) ;
			swap(b[x] , b[y]) ;
			B.ins(x , b[x]) ; B.ins(y , b[y]) ;
		}
	}
	return out.flush(), 0;
}