【BZOJ 4103】【THUSC 2015】异或运算

http://www.lydsy.com/JudgeOnline/problem.php?id=4103
对长的那一维建可持久化trie树（主席树？）
最主要的思路是对短的那一维每一位暴力，每一位都记录分别匹配到了trie上的哪两个点（区间左开右闭，所以两个点）。
时间复杂度(O(nplog m))。

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

const int N = 300003;

struct node {int ch[2], s;} T[N * 32];
int cnt = 0, root[N], x[1003], y[N], n, m, tl[1003], tr[1003];

void update(int &pos, int num, int tmp) {
	T[++cnt] = T[pos]; pos = cnt; ++T[pos].s;
	if (tmp == -1) return;
	update(T[pos].ch[(num >> tmp) & 1], num, tmp - 1);
}

int query(int u, int d, int k, int tmp) {
	if (tmp == -1) return 0;
	if (tmp == 2) {
		++tmp; --tmp;
	}
	int s = 0, v;
	for (int i = u; i <= d; ++i) {
		v = ((x[i] >> tmp) & 1) ^ 1;
		s += T[T[tr[i]].ch[v]].s - T[T[tl[i]].ch[v]].s;
	}
	if (s >= k) {
		for (int i = u; i <= d; ++i) {
			v = ((x[i] >> tmp) & 1) ^ 1;
			tl[i] = T[tl[i]].ch[v];
			tr[i] = T[tr[i]].ch[v];
		}
		return query(u, d, k, tmp - 1) | (1 << tmp);
	} else {
		k -= s;
		for (int i = u; i <= d; ++i) {
			v = (x[i] >> tmp) & 1;
			tl[i] = T[tl[i]].ch[v];
			tr[i] = T[tr[i]].ch[v];
		}
		return query(u, d, k, tmp - 1);
	}
}

int main() {
	scanf("%d%d", &n, &m);
	for (int i = 1; i <= n; ++i) scanf("%d", x + i);
	for (int i = 1; i <= m; ++i) {
		root[i] = root[i - 1];
		scanf("%d", y + i);
		update(root[i], y[i], 30);
	}
	
	int p, u, d, l, r, k; scanf("%d", &p);
	while (p--) {
		scanf("%d%d%d%d%d", &u, &d, &l, &r, &k);
		for (int i = u; i <= d; ++i)
			tl[i] = root[l - 1], tr[i] = root[r];
		printf("%d
", query(u, d, k, 30));
	}
	return 0;
}