[SCOI2016]幸运数字

嘟嘟嘟


首先都知道这题得用线性基。
然后预处理出线性基的倍增数组，查询的时候找lca的同时维护路径的线性基数组。
倍增lca复杂度(O(logn))，合并线性基(O(log ^ 2 n))，所以总复杂度(O(n log ^ 3 n))。
注意的是，查询的时候如果一步都跳不了，(x)和(y)所在点的权值就没有加入线性基。所以在跳之前就先把(x)和(y)加入。重了也没有关系。

#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<vector>
#include<stack>
#include<queue>
using namespace std;
#define enter puts("") 
#define space putchar(' ')
#define Mem(a, x) memset(a, x, sizeof(a))
#define In inline
typedef long long ll;
typedef double db;
const int INF = 0x3f3f3f3f;
const db eps = 1e-8;
const int maxn = 2e4 + 5;
const int maxb = 15;
const int maxN = 60;
inline ll read()
{
	ll ans = 0;
	char ch = getchar(), last = ' ';
	while(!isdigit(ch)) {last = ch; ch = getchar();}
	while(isdigit(ch)) {ans = (ans << 1) + (ans << 3) + ch - '0'; ch = getchar();}
	if(last == '-') ans = -ans;
	return ans;
}
inline void write(ll x)
{
	if(x < 0) x = -x, putchar('-');
	if(x >= 10) write(x / 10);
	putchar(x % 10 + '0');
}

int n, q;
ll a[maxn];
struct Edge
{
	int nxt, to;
}e[maxn << 1];
int head[maxn], ecnt = -1;
In void addEdge(int x, int y)
{
	e[++ecnt] = (Edge){head[x], y};
	head[x] = ecnt;
}

int dep[maxn], fa[maxb + 5][maxn];
ll p[maxb + 5][maxn][maxN + 5];
In void insert(ll* p, ll x)
{
	for(int i = maxN; i >= 0; --i)
	{
		if((x >> i) & 1)
		{
			if(p[i]) x ^= p[i];
			else {p[i] = x; return;}
		}
	}
}
In void merge(ll* A, ll* B, ll* C)
{
	for(int i = 0; i <= maxN; ++i) C[i] = A[i];
	for(int i = 0; i <= maxN; ++i) if(B[i]) insert(C, B[i]);
}
In void dfs(int now, int _f)
{
	insert(p[0][now], a[now]);
	for(int i = 1; (1 << i) <= dep[now]; ++i)
	{
		fa[i][now] = fa[i - 1][fa[i - 1][now]];
		merge(p[i - 1][now], p[i - 1][fa[i - 1][now]], p[i][now]);
	}
	for(int i = head[now], v; i != -1; i = e[i].nxt)
	{
		if((v = e[i].to) == _f) continue;
		dep[v] = dep[now] + 1;
		fa[0][v] = now;
		insert(p[0][v], a[now]);
		dfs(v, now);
	}
}
ll ans[maxN + 5];
In void merge2(ll* ret, ll* a)
{
	for(int i = 0; i <= maxN; ++i) if(a[i]) insert(ret, a[i]);
}
In ll query(int x, int y)
{
	Mem(ans, 0); insert(ans, a[x]); insert(ans, a[y]);
	if(dep[x] < dep[y]) swap(x, y);
	for(int i = maxb; i >= 0; --i) if(dep[x] - (1 << i) >= dep[y])
	{
		merge2(ans, p[i][x]);
		x = fa[i][x];
	}
	if(x ^ y)
	{
		for(int i = maxb; i >= 0; --i) if(fa[i][x] ^ fa[i][y])
		{
			merge2(ans, p[i][x]);
			merge2(ans, p[i][y]);
			x = fa[i][x]; y = fa[i][y];
		}
		insert(ans, a[fa[0][x]]);
	}
	ll ret = 0;
	for(int i = maxN; i >= 0; --i) if(!((ret >> i) & 1)) ret ^= ans[i];
	return ret;
}

int main()
{
	Mem(head, -1);
	n = read(); q = read();
	for(int i = 1; i <= n; ++i) a[i] = read();
	for(int i = 1; i < n; ++i)
	{
		int x = read(), y = read();
		addEdge(x, y); addEdge(y, x);
	}
	dfs(1, 0);
	for(int i = 1; i <= q; ++i)
	{
		int x = read(), y = read();
		write(query(x, y)), enter;
	}
	return 0;
}