2019.03.25 bzoj4568: [Scoi2016]幸运数字（倍增+线性基）

传送门

题意：给你一棵带点权的树，多次询问路径的最大异或和。

---

思路：

线性基上树？？

倍增维护一下就完了。

时间复杂度O(nlog3n)O(nlog^3n)O(nlog3n)

代码：

#include<bits/stdc++.h>
#define ri register int
#define fi first
#define se second
using namespace std;
typedef long long ll;
const int rlen=1<<18|1;
inline char gc(){
    static char buf[rlen],*ib,*ob;
    (ib==ob)&&(ob=(ib=buf)+fread(buf,1,rlen,stdin));
    return ib==ob?-1:*ib++;
}
inline ll read(){
    ll ans=0;
    char ch=gc();
    while(!isdigit(ch))ch=gc();
    while(isdigit(ch))ans=((ans<<2)+ans<<1)+(ch^48),ch=gc();
    return ans;
}
const int N=20005;
int n,m,fa[N][15],dep[N];
ll a[N];
struct xxj{
    ll a[61];
    xxj(){memset(a,0,sizeof(a));}
    inline void insert(ll x){
        for(ri i=60;~i;--i)if((x>>i)&1){
            if(a[i]){x^=a[i];continue;}
            a[i]=x;
            break;
        }
    }
    friend inline xxj operator+(const xxj&a,const xxj&b){
        xxj ret=a;
        for(ri i=60;~i;--i)if(b.a[i])ret.insert(b.a[i]);
        return ret;
    }
    inline ll query(){
        ll ret=0;
        for(ri i=60;~i;--i)ret=max(ret,ret^a[i]);
        return ret;
    }
}st[N][15];
vector<int>e[N];
void dfs(int p){
    st[p][0].insert(a[fa[p][0]]);
    for(ri i=0,v;i<e[p].size();++i){
        if((v=e[p][i])==fa[p][0])continue;
        fa[v][0]=p,dep[v]=dep[p]+1,dfs(v);
    }
}
inline ll solve(int x,int y){
    xxj ret;
    ret.insert(a[x]),ret.insert(a[y]);
    if(dep[x]<dep[y])swap(x,y);
    for(ri i=14,tmp=dep[x]-dep[y];~i;--i)if((tmp>>i)&1)ret=ret+st[x][i],x=fa[x][i];
    if(x==y)return ret.query();
    for(ri i=14;~i;--i)if(fa[x][i]^fa[y][i])ret=ret+(st[x][i]+st[y][i]),x=fa[x][i],y=fa[y][i];
    return (ret+st[x][0]).query();
}
int main(){
    n=read(),m=read();
    for(ri i=1;i<=n;++i)a[i]=read();
    for(ri i=1,u,v;i<n;++i)u=read(),v=read(),e[u].push_back(v),e[v].push_back(u);
    dfs(1);
    for(ri j=1;j<15;++j)for(ri i=1;i<=n;++i)fa[i][j]=fa[fa[i][j-1]][j-1],st[i][j]=st[i][j-1]+st[fa[i][j-1]][j-1];
    for(ri x,y;m;--m)x=read(),y=read(),cout<<solve(x,y)<<'\n';
    return 0;
}