BZOJ 4568 [Scoi2016]幸运数字（树链剖分 + 异或线性基）

题目链接  BZOJ 4568

考虑树链剖分+线段树维护每一段区域的异或线性基

对于每个询问，求出该点集的异或线性基。然后求一下这个线性基里面能异或出的最大值即可。

#include <bits/stdc++.h>

using namespace std;

#define rep(i, a, b)	for (int i(a); i <= (b); ++i)
#define dec(i, a, b)	for (int i(a); i >= (b); --i)
#define ls		i << 1
#define	rs		i << 1 | 1
#define lson		i << 1, L, mid
#define rson		i << 1 | 1, mid + 1, R

typedef long long LL;

const int N = 2e4 + 10;

struct lb{
	LL d[70];
	int cnt;
	void clear(){
		memset(d, 0, sizeof d);
		cnt = 0;
	}
	bool ins(LL val){
		dec(i, 62, 0) if (val & (1LL << i)){
			if (!d[i]){ d[i] = val; break; }
			val ^= d[i];
		}
		return val > 0;
	}
	LL qmax(){
		LL ret = 0;
		dec(i, 62, 0) if ((ret ^ d[i]) > ret) ret ^= d[i];
		return ret;
	}
	LL qmin(){
		rep(i, 0, 62) if (d[i]) return d[i];
		return 0;
	}
};

lb t[N << 3];

int f[N], fp[N], son[N], deep[N], father[N], sz[N], top[N];
int q, tot, n;
LL a[N];
vector <int> v[N];

lb merge(const lb &n1, const lb &n2){
	lb ret = n1;
	dec(i, 60, 0) if (n2.d[i]) ret.ins(n2.d[i]);
	return ret;
}

void dfs1(int x, int fa, int dep){
	deep[x] = dep;
	father[x] = fa;
	son[x] = 0;
	sz[x] = 1;
	int ct = (int)v[x].size();
	rep(i, 0, ct - 1){
		int u = v[x][i];
		if (u == fa) continue;
		dfs1(u, x, dep + 1);
		sz[x] += sz[u];
		if (sz[son[x]] < sz[u]) son[x] = u;
	}
}

void dfs2(int x, int tp){
	top[x] = tp;
	f[x] = ++tot;
	fp[f[x]] = x;
	if (son[x]) dfs2(son[x], tp);
	int ct = (int)v[x].size();
	rep(i, 0, ct - 1){
		int u = v[x][i];
		if (u == father[x] || u == son[x]) continue;
		dfs2(u, u);
	}
}

void build(int i, int L, int R){
	if (L == R){
		t[i].ins(a[fp[L]]);
		return;
	}

	int mid = (L + R) >> 1;
	build(lson);
	build(rson);
	t[i] = merge(t[ls], t[rs]);
}

lb query(int i, int L, int R, int l, int r){
	if (L == l && R == r) return t[i];
	int mid = (L + R) >> 1;
	if (r <= mid) return query(lson, l, r);
	else if (l > mid) return query(rson, l, r);
	else return merge(query(lson, l, mid), query(rson, mid + 1, r));
}

lb work(int x, int y){
	lb ret;
	ret.clear();
	int f1 = top[x], f2 = top[y];
	for (; f1 != f2; ){
		if (deep[f1] < deep[f2]) swap(f1, f2), swap(x, y);
		ret = merge(ret, query(1, 1, n, f[f1], f[x]));
		x = father[f1], f1 = top[x];
	}

	if (x == y) return merge(ret, query(1, 1, n, f[x], f[y]));
	if (deep[x] > deep[y]) swap(x, y);
	return merge(ret, query(1, 1, n, f[x], f[y]));
}

int main(){

	scanf("%d%d", &n, &q);
	rep(i, 1, n) scanf("%lld", a + i);

	rep(i, 2, n){
		int x, y;
		scanf("%d%d", &x, &y);
		v[x].push_back(y);
		v[y].push_back(x);
	}

	dfs1(1, 0, 0);
	dfs2(1, 1);
	build(1, 1, n);

	while (q--){
		int x, y;
		scanf("%d%d", &x, &y);
		lb cnt = work(x, y);
		printf("%lld\n", cnt.qmax());
	}

	return 0;
}