【HDOJ6579】Operation（线性基）

题意：给定一个数列a，给定两种操作：

1.询问[l,r]区间内最大的xor和

2.n++，a[n]赋值为x

要求强制在线

n,m<=5e5，a[i]<2^30

思路：同CF1100F

固定右端点，维护每一维上使生成空间变大的最大的左端点

 #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  1100000
 #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(),m=read();
         rep(i,,n)
          rep(j,,) f[i][j]=g[i][j]=;
         int lastans=;
         rep(i,,n)
         {
             a[i]=read();
             add(i,a[i]);
         }

         while(m--)
         {
             int op=read();
             if(op)
             {
                 a[++n]=read()^lastans;
                 add(n,a[n]);
             }
              else
              {
                  int l=read(),r=read();
                  l=(l^lastans)%n+,r=(r^lastans)%n+;
                  if(l>r) swap(l,r);
                  lastans=;
                  per(i,,)
                   if((lastans^f[r][i])>lastans&&g[r][i]>=l) lastans^=f[r][i];
                  printf("%d\n",lastans);
              }
         }

     }

     return ;
 }