这题其实很奇怪,一眼看出fhq treap+字符串hash。。。
结果在BZOJ被卡常了,结果学长告诉我了一些优化,有些自己还不会写,只写了inline和吧unsigned long long改为unsigned int。。。
结果时间少了一半,谁能告诉我为什么。。。
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
using namespace std;
int m,len;
char s[100000];
unsigned int mi[200000];
struct treap_node{
treap_node *left,*right;
int fix,size;
unsigned int hashz;
char val;
treap_node(char val): val(val) {left=right=NULL; fix=rand(); hashz=(int)(val)-96; size=1; }
int lsize()
{
if (left)
return left->size;
else
return 0;
}
int rsize()
{
if (right)
return right->size;
else
return 0;
}
void Maintain()
{
size=1;
size+=lsize()+rsize();
hashz=(int)(val)-96;
if (left&&!right)
hashz=left->hashz+hashz*mi[lsize()];
if (!left&&right)
hashz=hashz+right->hashz*mi[1];
if (left&&right)
hashz=left->hashz+hashz*mi[lsize()]+right->hashz*mi[lsize()+1];
}
};
treap_node *root;
typedef pair<treap_node*,treap_node*> droot;
inline treap_node *merge(treap_node *a,treap_node *b)
{
if (!a) return b; if (!b) return a;
if (a->fix<b->fix)
{
a->right=merge(a->right,b);
a->Maintain();
return a;
}
else
{
b->left=merge(a,b->left);
b->Maintain();
return b;
}
}
inline droot split(treap_node *x,int k)
{
if (!x) return droot(NULL,NULL);
droot y;
if (k<=x->lsize())
{
y=split(x->left,k);
x->left=y.second;
x->Maintain();
y.second=x;
}
else
{
y=split(x->right,k-x->lsize()-1);
x->right=y.first;
x->Maintain();
y.first=x;
}
return y;
}
inline void insert(int xi,char ch)
{
droot x;
treap_node *temp;
x=split(root,xi);
temp=new treap_node(ch);
x.first=merge(x.first,temp);
root=merge(x.first,x.second);
}
inline void del(int xi,int k)
{
droot x,y;
x=split(root,xi-1);
y=split(x.second,1);
root=merge(x.first,y.second);
}
inline void work(int x,int y)
{
int l,r,mid,ans=0;
unsigned int t1,t2;
droot a,b;
l=1; r=min(len-x+1,len-y+1);
while (l<=r)
{
mid=(l+r)/2;
a=split(root,x-1); b=split(a.second,mid);
b.first->Maintain();
t1=b.first->hashz;
a.second=merge(b.first,b.second); root=merge(a.first,a.second);
a=split(root,y-1); b=split(a.second,mid);
b.first->Maintain();
t2=b.first->hashz;
a.second=merge(b.first,b.second); root=merge(a.first,a.second);
if (t1==t2)
l=mid+1,ans=mid;
else
r=mid-1;
}
printf("%d\n",ans);
}
int main()
{
int i,x,y,t=0;
char opt,ch;
treap_node *temp;
scanf("%s",&s);
len=strlen(s);
scanf("%d",&m);
mi[0]=1;
for (i=1;i<=100000;++i)
mi[i]=mi[i-1]*27;
for (i=0;i<=len-1;++i)
{
temp=new treap_node(s[i]);
root=merge(root,temp);
}
for (i=1;i<=m;++i)
{
opt=getchar();
while (opt!=‘Q‘&&opt!=‘R‘&&opt!=‘I‘)
opt=getchar();
if (opt==‘Q‘)
{
scanf("%d%d",&x,&y);
work(x,y);
}
if (opt==‘I‘)
{
scanf("%d",&x);
ch=getchar();
while (ch<‘a‘||ch>‘z‘)
ch=getchar();
insert(x,ch);
len++;
}
if (opt==‘R‘)
{
scanf("%d",&x);
ch=getchar();
while (ch<‘a‘||ch>‘z‘)
ch=getchar();
del(x,1);
insert(x-1,ch);
}
}
}
版权声明:本文为博主原创文章,未经博主允许不得转载。
时间: 2024-10-10 02:15:17