BZOJ.4184.shallot(线段树分治 线性基)
裸的线段树分治+线性基,就是跑的巨慢_(:з」∠)_ 。
不知道他们都写的什么=-=
//41652kb 11920ms
#include <map>
#include <cstdio>
#include <cctype>
#include <vector>
#include <algorithm>
#define BIT 30
#define gc() getchar()
#define MAXIN 500000
//#define gc() (SS==TT&&(TT=(SS=IN)+fread(IN,1,MAXIN,stdin),SS==TT)?EOF:*SS++)
typedef long long LL;
const int N=5e5+5;
std::map<int,int> las;
char IN[MAXIN],*SS=IN,*TT=IN;
struct Base
{
int x[BIT+1];
void Insert(int v)
{
for(int i=BIT; ~i; --i)
if(v>>i&1)
if(x[i]) v^=x[i];
else {x[i]=v; break;}
}
void Query()
{
int res=0;
for(int i=BIT; ~i; --i) res=std::max(res,res^x[i]);
printf("%d\n",res);
}
}base;
struct Segment_Tree
{
#define ls rt<<1
#define rs rt<<1|1
#define lson l,m,ls
#define rson m+1,r,rs
#define S N<<2
std::vector<int> vec[S];
#undef S
void Modify(int l,int r,int rt,int L,int R,int v)
{
if(L<=l && r<=R) {vec[rt].push_back(v); return;}
int m=l+r>>1;
if(L<=m) Modify(lson,L,R,v);
if(m<R) Modify(rson,L,R,v);
}
void Solve(int l,int r,int rt,Base b)
{
for(std::vector<int>::iterator it=vec[rt].begin(); it!=vec[rt].end(); ++it) b.Insert(*it);
if(l==r) {b.Query(); return;}
int m=l+r>>1; Solve(lson,b), Solve(rson,b);
}
}T;
inline int read()
{
int now=0;register char c=gc();
for(;!isdigit(c);c=gc());
for(;isdigit(c);now=now*10+c-48,c=gc());
return now;
}
int main()
{
#define S 1,n,1
const int n=read();
for(int i=1; i<=n; ++i)
{
int a=read();
if(las[a]) T.Modify(S,las[a],i-1,a), las[a]=0;
else las[a]=i;
}
for(std::map<int,int>::iterator it=las.begin(); it!=las.end(); ++it)
if((*it).second) T.Modify(S,(*it).second,n,(*it).first);
T.Solve(S,base);
return 0;
}
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很久以前的奇怪但现在依旧成立的签名
attack is our red sun $$\color{red}{\boxed{\color{red}{attack\ is\ our\ red\ sun}}}$$ ------------------------------------------------------------------------------------------------------------------------
很久以前的奇怪但现在依旧成立的签名
attack is our red sun $$\color{red}{\boxed{\color{red}{attack\ is\ our\ red\ sun}}}$$ ------------------------------------------------------------------------------------------------------------------------