BZOJ 4546: codechef XRQRS

可持久化Trie本质和主席树一致
但能支持求最大异或和操作
抛开这个功能测过可持久化trie去写主席树的题，好像也没快多少？可能是我写丑了

#include <bits/stdc++.h>
#define pb push_back
#define fi first
#define se second
#define pii pair<int, int>
#define pli pair<ll, int>
#define lp p << 1
#define rp p << 1 | 1
#define mid ((l + r) / 2)
#define lowbit(i) ((i) & (-i))
typedef long long ll;
typedef unsigned long long ull;
typedef double db;
#define rep(i,a,b) for(int i=a;i<b;i++)
#define per(i,a,b) for(int i=b-1;i>=a;i--)
#define Edg int ccnt=1,head[N],to[E],ne[E];void addd(int u,int v){to[++ccnt]=v;ne[ccnt]=head[u];head[u]=ccnt;}void add(int u,int v){addd(u,v);addd(v,u);}
#define Edgc int ccnt=1,head[N],to[E],ne[E],c[E];void addd(int u,int v,int w){to[++ccnt]=v;ne[ccnt]=head[u];c[ccnt]=w;head[u]=ccnt;}void add(int u,int v,int w){addd(u,v,w);addd(v,u,w);}
#define es(u,i,v) for(int i=head[u],v=to[i];i;i=ne[i],v=to[i])
const int MOD = 1e9 + 7;
void M(int &x) {if (x >= MOD)x -= MOD; if (x < 0)x += MOD;}
int qp(int a, int b = MOD - 2) {int ans = 1; for (; b; a = 1LL * a * a % MOD, b >>= 1)if (b & 1)ans = 1LL * ans * a % MOD; return ans % MOD;}
template<class T>T gcd(T a, T b) { while (b) { a %= b; std::swap(a, b); } return a; }
template<class T>bool chkmin(T &a, T b) { return a > b ? a = b, 1 : 0; }
template<class T>bool chkmax(T &a, T b) { return a < b ? a = b, 1 : 0; }
char buf[1 << 21], *p1 = buf, *p2 = buf;
inline char getc() {
	return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2) ? EOF : *p1++;
}
inline int _() {
	int x = 0, f = 1; char ch = getc();
	while (ch < '0' || ch > '9') { if (ch == '-') f = -1; ch = getc(); }
	while (ch >= '0' && ch <= '9') { x = x * 10ll + ch - 48; ch = getc(); }
	return x * f;
}

const int N = 500011;
int sz[N * 21], ch[N * 21][2], tol, root[N], n;
int ins(int u, int x) {
	int p = ++tol, p0 = p;
	per (i, 0, 20) {
		int d = x >> i & 1;
		ch[p][d ^ 1] = ch[u][d ^ 1];
		sz[p = ch[p][d] = ++tol] = sz[u = ch[u][d]] + 1;
	}
	return p0;
}
int qmx(int p, int q, int x) {
	int ans = 0;
	per (i, 0, 20) {
		int d = x >> i & 1;
		if (sz[ch[p][d ^ 1]] - sz[ch[q][d ^ 1]] > 0)
			p = ch[p][d ^ 1], q = ch[q][d ^ 1], ans += (1 << i);
		else 
			p = ch[p][d], q = ch[q][d];
	}
	return ans ^ x;
}
int qkth(int p, int q, int k) {
	int res = 0;
	per (i, 0, 20) {
		int t = sz[ch[p][0]] - sz[ch[q][0]];
		if (t >= k) p = ch[p][0], q = ch[q][0];
		else k -= t, p = ch[p][1], q = ch[q][1], res += (1 << i);
	}
	return res;
}
int qcnt(int p, int q, int x) {
	x++;
	int ans = 0;
	per (i, 0, 20) {
		int d = x >> i & 1;
		if (d) ans += sz[ch[p][0]] - sz[ch[q][0]];
		p = ch[p][d], q = ch[q][d];
	}
	return ans;
}

int main() {
#ifdef LOCAL
	freopen("ans.out", "w", stdout);
#endif
	int q = _();
	while (q--) {
		int opt = _();
		if (opt == 1) {
			int x = _();
			n++;
			root[n] = ins(root[n - 1], x);
		} else if (opt == 2) {
			int l = _(), r = _(), x = _();
			printf("%d\n", qmx(root[r], root[l - 1], x));
		} else if (opt == 3) {
			int k = _();
			n -= k;
		} else if (opt == 4) {
			int l = _(), r = _(), x = _();
			printf("%d\n", qcnt(root[r], root[l - 1], x));
		} else {
			int l = _(), r = _(), x = _();
			printf("%d\n", qkth(root[r], root[l - 1], x));
		}
	}
	return 0;
}