_AyachiNene @ 2024-11-28 21:22:56
rt,这个题 我用cchash和umap都能过但是用gphash #5直接满了几百倍,求大佬解答。
#include<bits/stdc++.h>
#include<ext/pb_ds/assoc_container.hpp>
#define ll unsigned long long
using namespace std;
using namespace __gnu_pbds;
namespace IO
{
char buff[1<<21],*p1=buff,*p2=buff;
char getch(){return p1==p2&&(p2=((p1=buff)+fread(buff,1,1<<21,stdin)),p1==p2)?EOF:*p1++;}
template<typename T>void read(T &x){char ch=getch();int fl=1;x=0;while(ch>'9'||ch<'0'){if(ch=='-')fl=-1;ch=getch();}while(ch<='9'&&ch>='0'){x=x*10+ch-48;ch=getch();}x*=fl;}
template<typename T>void read_s(T &x){char ch=getch();while(ch<'a'||ch>'z')ch=getch();while(ch>='a'&&ch<='z'){x+=ch;ch=getch();}}
template<typename T,typename ...Args>void read(T &x,Args& ...args){read(x);read(args...);}
char obuf[1<<21],*p3=obuf;
void putch(char ch) {if(p3-obuf<(1<<21))*p3++=ch;else fwrite(obuf,p3-obuf,1,stdout),p3=obuf,*p3++=ch;}
char ch[100];
template<typename T>void write(T x) {if(!x)return putch('0');if(x<0)putch('-'),x*=-1;int top=0;while(x)ch[++top]=x%10+48,x/=10;while(top)putch(ch[top]),top--;}
template<typename T,typename ...Args>void write(T x,Args ...args) {write(x);write(args...);}
void put(string s){for(int i=0;i<s.size();i++)putch(s[i]);}
void flush(){fwrite(obuf,p3-obuf,1,stdout);}
}
using namespace IO;
int n,m,q;
vector<int>e[500005];
ll Pow[500005];
ll a[500005];
int f[500005];
ll h[500005];
const ll base=(2e6+3);
cc_hash_table<ll,int>mp;
//unordered_map<ll,int>mp;
void dfs(int u)
{
mp[h[u]]=u;
for(int i=0;i<e[u].size();i++)
{
int v=e[u][i];
h[v]=h[u]*base+i+1;
dfs(v);
}
}
namespace Nene
{
struct segt
{
int len;
ll val;
inline segt operator+(const segt &a)const
{
segt res;
res.len=len+a.len;
res.val=val*Pow[a.len]+a.val;
return res;
}
}t[500005<<2];
#define ls (root<<1)
#define rs (root<<1|1)
#define mid (l+r>>1)
void insert(int x,ll v,int root=1,int l=1,int r=m)
{
if(l==r) return t[root].val=v,t[root].len=1,void();
if(x<=mid) insert(x,v,ls,l,mid);
else insert(x,v,rs,mid+1,r);
t[root]=t[ls]+t[rs];
}
vector<int>rt(100),tl(100),tr(100);
void get(int x,int y,int root=1,int l=1,int r=m)
{
if(l>=x&&r<=y) return rt.push_back(root),tl.push_back(l),tr.push_back(r),void();
if(x<=mid) get(x,y,ls,l,mid);
if(y>mid) get(x,y,rs,mid+1,r);
}
const ll f1=18446744073709551615ull;
ll find(ll now,int root,int l,int r)
{
if(mp.find(now)==mp.end()) return f1;
if(l==r) return mp.find(now*base+t[root].val)==mp.end()?now:now*base+t[root].val;
if(mp.find(now*Pow[t[ls].len]+t[ls].val)==mp.end()) return find(now,ls,l,mid);
return find(now*Pow[t[ls].len]+t[ls].val,rs,mid+1,r);
}
inline ll query(int p,int x,int y)
{
rt.clear();tl.clear();tr.clear();
ll now=h[p];
get(x,y);
for(int i=0;i<rt.size();i++)
{
if(mp.find(now*Pow[t[rt[i]].len]+t[rt[i]].val)==mp.end())
{
ll tmp=find(now,rt[i],tl[i],tr[i]);
return tmp==f1?p:mp[tmp];
}
now=now*Pow[t[rt[i]].len]+t[rt[i]].val;
}
return mp[now];
}
#undef mid
}using namespace Nene;
int main()
{
// freopen("P5537_5.in","r",stdin);
// freopen(".out","w",stdout);
read(n,m,q);
int rt;
for(int i=1;i<=n;i++)
{
read(f[i]);
if(!f[i]) rt=i;
else e[f[i]].push_back(i);
}
for(int i=1;i<=n;i++) sort(e[i].begin(),e[i].end());
Pow[0]=1;
for(int i=1;i<=500000;i++) Pow[i]=Pow[i-1]*base;
dfs(rt);
for(int i=1;i<=m;i++) read(a[i]),insert(i,a[i]);
while(q--)
{
int op,x,ql,qr;read(op,x,ql);
if(op==1) read(qr),write(query(x,ql,qr)),putch('\n');
else insert(x,ql),a[x]=ql;
}
flush();
return 0;
}
by Little_Cart @ 2024-11-28 21:27:55
孩子们,gp和cc都可以被卡成
by Little_Cart @ 2024-11-28 21:28:26
@_AyachiNene
by Miss_SGT @ 2024-11-28 21:29:42
by _AyachiNene @ 2024-11-28 21:29:50
@Little_Cart 还以为卡不了,thx
by Miss_SGT @ 2024-11-28 21:30:13
自己手写最快
by qazsedcrfvgyhnujijn @ 2024-11-28 21:30:28
理论上探查比拉链快吧。
试试显式重载一个基于时间的随机数种子,这样应该真的就卡不了了吧。
实在不行上类似双模的两个 hash_table
?