【CF1100F】Ivan and Burgers（线性基，分治）

题意：给定n个数，每个数为c[i]，有q个询问，每次询问从第l个到第r个数字的最大xor和

n,q<=5e5，c[i]<=1e6，时限3s

思路：直接线段树维护区间线性基是3个log，会T

做法1：因为不是强制在线把询问分治能降到2个log

 #include<bits/stdc++.h>
 #include<cstdio>
 #include<cstring>
 #include<string>
 #include<cmath>
 #include<iostream>
 #include<algorithm>
 #include<map>
 #include<set>
 #include<queue>
 #include<vector>
 #include<bitset>
 using namespace std;
 typedef long long ll;
 typedef unsigned int uint;
 typedef unsigned long long ull;
 typedef pair<int,int> PII;
 typedef vector<int> VI;
 typedef set<int>::iterator iter;
 #define fi first
 #define se second
 #define MP make_pair
 #define mem0(a) memset(a,0,sizeof(a))
 #define N      510000
 #define M      15
 #define MOD 1000000007
 #define eps 1e-10
 #define pi     acos(-1)
 #define oo     1e9
  
 int read()
 {
    int v=,f=;
    char c=getchar();
    while(c<||<c) {if(c=='-') f=-; c=getchar();}
    while(<=c&&c<=) v=(v<<)+v+v+c-,c=getchar();
    return v*f;
 }
  
 struct base
 {
     //ll d[20],p[20];
     int d[];
     int cnt;

     base()
     {
         memset(d,,sizeof(d));
         //memset(p,0,sizeof(p));
         cnt=;
     }
     bool insert(int val)
     {
         if(val==) return ;
         for(int i=;i>=;i--)
          if((val>>i)&)
             {
                 if(!d[i])
                 {
                     d[i]=val;
                     break;
                 }
                 val^=d[i];
             }
         return val>;
     }
     int query_max()
     {
         int ret=;
         for(int i=;i>=;i--) ret=max(ret,ret^d[i]);
         return ret;
     }
     int query_min()
     {
         for(int i=;i<=;i++)
          if(d[i]) return d[i];
         return ;
     }
     /*void rebuild()
     {
         for(int i=20;i>=0;i--)
          for(int j=i-1;j>=0;j--)
           if((d[i]>>j)&1) d[i]^=d[j];
         for(int i=0;i<=20;i++)
          if(d[i]) p[cnt++]=d[i];
     }
     ll kthquery(ll k)
     {
         int ret=0;
         if(k>=(1LL<<cnt)) return -1;
         for(int i=20;i>=0;i--)
          if((k>>i)&1) ret^=p[i];
         return ret;
     }*/

     base operator+(const base&_A)const
     {
         base ret=*this;
         for(int i=;i>=;i--)
          if(_A.d[i]) ret.insert(_A.d[i]);
         return ret;
     }
 }t[N],b;
  
 struct arr
 {
     int x,y,id;
 }q[N],c[N];
  
 int a[N],ans[N];

 /*base merge(const base &n1,const base &n2)
 {
     base ret=n1;
     for(int i=20;i>=0;i--)
      if(n2.d[i]) ret.insert(n1.d[i]);
     return ret;
 }*/
  
 void solve(int l,int r,int x,int y)
 {
     if(l==r)
     {
         for(int i=x;i<=y;i++) ans[q[i].id]=a[l];
         return;
     }
     if(x>y) return;
     int mid=(l+r)>>;
     memset(b.d,,sizeof(b.d));
     for(int i=mid;i>=l;i--)
     {
         b.insert(a[i]);
         for(int j=;j<=;j++) t[i].d[j]=b.d[j];
     }
     memset(b.d,,sizeof(b.d));
     for(int i=mid+;i<=r;i++)
     {
         b.insert(a[i]);
         for(int j=;j<=;j++) t[i].d[j]=b.d[j];
     }
     int l1=x,r1=y;
     for(int i=x;i<=y;i++)
     {
         if(q[i].y<=mid) c[l1++]=q[i];
          else if(q[i].x>mid) c[r1--]=q[i];
           else
           {
                   base tmp=t[q[i].x]+t[q[i].y];
                 ans[q[i].id]=tmp.query_max();
           }
     }
     for(int i=x;i<l1;i++) q[i]=c[i];
     for(int i=r1+;i<=y;i++) q[i]=c[i];
     solve(l,mid,x,l1-);
     solve(mid+,r,r1+,y);
 }
  
 int main()
 {
     //freopen("cf1100f.in","r",stdin);
     //freopen("cf1100f.out","w",stdout);
     int n,m;
     scanf("%d",&n);
     for(int i=;i<=n;i++) scanf("%d",&a[i]);
     scanf("%d",&m);
     for(int i=;i<=m;i++)
     {
         scanf("%d%d",&q[i].x,&q[i].y);
         q[i].id=i;
     }
     solve(,n,,m);
     for(int i=;i<=m;i++) printf("%d\n",ans[i]);
     return ;
 }

做法2：假设固定右端点，维护log个维上使生成空间变大的最大的l

比做法1快3倍

 #include<bits/stdc++.h>
 using namespace std;
 typedef long long ll;
 typedef unsigned int uint;
 typedef unsigned long long ull;
 typedef pair<int,int> PII;
 typedef pair<ll,ll> Pll;
 typedef vector<int> VI;
 typedef vector<PII> VII;
 typedef pair<ll,int>P;
 #define N  510000
 #define M  151000
 #define fi first
 #define se second
 #define MP make_pair
 #define pi acos(-1)
 #define mem(a,b) memset(a,b,sizeof(a))
 #define rep(i,a,b) for(int i=(int)a;i<=(int)b;i++)
 #define per(i,a,b) for(int i=(int)a;i>=(int)b;i--)
 #define lowbit(x) x&(-x)
 #define Rand (rand()*(1<<16)+rand())
 #define id(x) ((x)<=B?(x):m-n/(x)+1)
 #define ls p<<1
 #define rs p<<1|1
  
 const ll MOD=1e9+,inv2=(MOD+)/;
       double eps=1e-;
       ll INF=1e18;
       ll inf=5e13;
       int dx[]={-,,,};
       int dy[]={,,-,};
  
 int f[N][],g[N][],a[N];
  
 int read()
 {
    int v=,f=;
    char c=getchar();
    while(c<||<c) {if(c=='-') f=-; c=getchar();}
    while(<=c&&c<=) v=(v<<)+v+v+c-,c=getchar();
    return v*f;
 }
  
 void add(int i,int x)
 {
     int k=i;
     per(j,,) f[i][j]=f[i-][j],g[i][j]=g[i-][j];
     per(j,,)
      if(x>>j)
      {
          if(!f[i][j])
          {
              f[i][j]=x;
              g[i][j]=k;
              break;
          }
           else
           {
               if(k>g[i][j])
               {
                   swap(k,g[i][j]);
                   swap(x,f[i][j]);
               }
               x^=f[i][j];
           }
      }
 }
  
 int main()
 {
     //freopen("1.in","r",stdin);
     //freopen("1.out","w",stdout);
     //int cas=read();
     //while(cas--)
     //{
         int n=read();
         rep(i,,n)
          rep(j,,) f[i][j]=g[i][j]=;
         int lastans=;
         rep(i,,n)
         {
             a[i]=read();
             add(i,a[i]);
         }
         int m=read();
         while(m--)
         {
             int l=read(),r=read();
             lastans=;
             per(i,,)
             if((lastans^f[r][i])>lastans&&g[r][i]>=l) lastans^=f[r][i];
             printf("%d\n",lastans);
         }
  
     //}
  
  
  
  
     return ;
 }