BZOJ4568 [Scoi2016]幸运数字 【点分治 + 线性基】

题目链接

BZOJ4568

题解

选任意个数异或和最大，使用线性基

线性基插入\(O(logn)\)，合并\(O(log^2n)\)

我们要求树上两点间异或和最大值，由于合并是\(O(log^2n)\)的，我们尽量只合并一次

那就采用点分治

每次求出到分治重心的线性基，将过分治重心的询问的两个线性基合并即可

复杂度\(O(60^2q + 60nlogn)\)

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<map>
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define mp(a,b) make_pair<int,int>(a,b)
#define cls(s) memset(s,0,sizeof(s))
#define cp pair<int,int>
#define LL long long int
using namespace std;
const int maxn = 20005,maxm = 200005,INF = 1000000000;
inline LL read(){
	LL out = 0,flag = 1; char c = getchar();
	while (c < 48 || c > 57){if (c == '-') flag = -1; c = getchar();}
	while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}
	return out * flag;
}
struct Bit{
	LL A[60];
	void init(){for (int i = 59; ~i; i--) A[i] = 0;}
	void copy(LL* B){for (int i = 59; ~i; i--) A[i] = B[i];}
	void ins(LL x){
		for (int i = 59; ~i; i--)
			if ((x >> i) & 1){
				if (A[i]) x ^= A[i];
				else {A[i] = x; break;}
			}
	}
	LL ask(){
		LL re = 0;
		for (int i = 59; ~i; i--) if ((re ^ A[i]) > re) re ^= A[i];
		return re;
	}
}B[maxn],T;
int h[maxn],ne = 1;
struct EDGE{int to,nxt;}ed[maxn << 1];
inline void build(int u,int v){
	ed[++ne] = (EDGE){v,h[u]}; h[u] = ne;
	ed[++ne] = (EDGE){u,h[v]}; h[v] = ne;
}
int g[maxn],nxt[30 * maxm],tq[30 * maxn],cnt;
int n,q,x[maxm],y[maxm],vis[maxn],pos[maxn],Vis[maxn],now;
LL G[maxn],ans[maxm];
void Add(int u,int v){
	nxt[++cnt] = g[u]; tq[cnt] = v; g[u] = cnt;
}
int F[maxn],siz[maxn],fa[maxn],N,rt;
void getrt(int u){
	F[u] = 0; siz[u] = 1;
	Redge(u) if (!vis[to = ed[k].to] && to != fa[u]){
		fa[to] = u; getrt(to);
		siz[u] += siz[to];
		F[u] = max(F[u],siz[to]);
	}
	F[u] = max(F[u],N - siz[u]);
	if (F[u] < F[rt]) rt = u;
}
void dfs(int u,int R){
	pos[u] = R; siz[u] = 1; Vis[u] = now;
	B[u].copy(B[fa[u]].A); B[u].ins(G[u]);
	Redge(u) if (!vis[to = ed[k].to] && to != fa[u]){
		fa[to] = u; dfs(to,R);
		siz[u] += siz[to];
	}
}
void solve(int u){
	F[rt = 0] = INF; N = siz[u]; getrt(u);
	//printf("u%d  rt%d\n",u,rt);
	pos[rt] = rt; vis[rt] = true; siz[u] = 1; Vis[rt] = ++now;
	B[rt].init(); B[rt].ins(G[rt]);
	Redge(rt) if (!vis[to = ed[k].to]){
		fa[to] = rt; dfs(to,to);
		siz[u] += siz[to];
	}
	for (int k = g[u],i,a,b; k; k = nxt[k]){
		i = tq[k]; a = x[i]; b = y[i];
		if (Vis[a] != now || Vis[b] != now) continue;
		if (pos[a] == pos[b]) Add(pos[a],i);
		else {
			T.copy(B[a].A);
			for (int j = 59; ~j; j--)
				if (B[b].A[j]) T.ins(B[b].A[j]);
			ans[i] = T.ask();
		}
	}
	Redge(rt) if (!vis[to = ed[k].to]){
		solve(to);
	}
}
int main(){
	n = read(); q = read();
	for (int i = 1; i <= n; i++) G[i] = read();
	for (int i = 1; i < n; i++) build(read(),read());
	for (int i = 1; i <= q; i++){
		x[i] = read(); y[i] = read();
		if (x[i] == y[i]) ans[i] = G[x[i]];
		else Add(1,i);
	}
	siz[1] = n; solve(1);
	for (int i = 1; i <= q; i++)
		printf("%lld\n",ans[i]);
	return 0;
}