BZOJ 4568: [Scoi2016]幸运数字 [线性基 倍增]

4568: [Scoi2016]幸运数字

题意：一颗带点权的树，求树上两点间异或值最大子集的异或值

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显然要用线性基

可以用倍增的思想，维护每个点向上\(2^j\)个祖先这些点的线性基，求lca的时候合并起来就行了

复杂度\(O(nlogn60*60)\)

注意这是点权，特判x==y的情况，需要插入a[x]

还可以用点分治和树链剖分

我的代码好慢啊...但是很好写啊

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
using namespace std;
typedef long long ll;
#define pii pair<int, int>
#define fir first
#define sec second
const int N=2e4+5;
inline ll read() {
    char c=getchar(); ll x=0, f=1;
    while(c<'0' || c>'9') {if(c=='-')f=-1; c=getchar();}
    while(c>='0' && c<='9') {x=x*10+c-'0'; c=getchar();}
    return x*f;
}
int n, Q, u, v; ll a[N];
struct edge{int v, ne;}e[N<<1];
int cnt=1, h[N];
inline void ins(int u, int v) {
	e[++cnt]=(edge){v, h[u]}; h[u]=cnt;
	e[++cnt]=(edge){u, h[v]}; h[v]=cnt;
}
namespace lb{
	struct meow {
		ll p[62];
		meow(){memset(p, 0, sizeof(p));}
		ll operator [](int x) {return p[x];}
		bool insert(ll val) {
			for(int i=60; i>=0; i--)
				if(val & (1LL<<i)) {
					if(p[i]) val ^= p[i];
					else {p[i]=val; break;}
				}
			return val>0;
		}
		void merge(meow &a) {
			for(int i=60; i>=0; i--) if(a[i]) insert(a[i]);
		}
		ll getmax() {
			ll ans=0;
			for(int i=60; i>=0; i--) ans = max(ans, ans^p[i]);
			return ans;
		}
		void print() {
			for(int i=60; i>=0; i--) if(p[i]) printf("p %d %lld\n",i,p[i]);
			puts("");
		}
	}b[N][17];
}using lb::b; using lb::meow;
int fa[N][17], deep[N];
void dfs(int u) {
	for(int i=1; (1<<i)<=deep[u]; i++) {
		fa[u][i] = fa[ fa[u][i-1] ][i-1];
		b[u][i].merge(b[u][i-1]); b[u][i].merge(b[ fa[u][i-1] ][i-1]);
	}
	for(int i=h[u];i;i=e[i].ne)
		if(e[i].v != fa[u][0]) {
			fa[e[i].v][0]=u;
			deep[e[i].v]=deep[u]+1;
			b[e[i].v][0].insert(a[u]);
			dfs(e[i].v);
		}
}
meow lca(int x, int y) { //printf("lca %d %d\n",x,y);
	meow now;
	if(x==y) {now.insert(a[x]); return now;}
	if(deep[x]<deep[y]) swap(x, y);
	int bin=deep[x]-deep[y];
	for(int i=0; i<15; i++)
		if((1<<i)&bin) now.merge(b[x][i]), x=fa[x][i];
	if(x==y) return now;
	for(int i=14; i>=0; i--)
		if(fa[x][i] != fa[y][i]) now.merge(b[x][i]), now.merge(b[y][i]), x=fa[x][i], y=fa[y][i];
	now.merge(b[x][0]); now.insert(a[y]);
	//puts("now");now.print();
	return now;
}
ll Que(int x, int y) {
	return lca(x, y).getmax();
}
int main() {
	freopen("in","r",stdin);
	n=read(); Q=read();
	for(int i=1; i<=n; i++) a[i]=read(), b[i][0].insert(a[i]);
	for(int i=1; i<n; i++) ins(read(), read());
	dfs(1);
	//for(int i=1; i<=n; i++)
	//	for(int j=0; (1<<j)<=deep[i]; j++) printf("hi %d  %d %d\n",i,j,fa[i][j]), b[i][j].print();
	for(int i=1; i<=Q; i++) printf("%lld\n", Que(read(), read()));
}