前言
寫寫比較麻煩的這題
題目相關
題目大意
寫一個大數據結構
數據範圍
20000
題目鏈接
前置
先過模板題,比如會個非旋treap,寫一下,通過[luogu3369][模板]普通平衡樹
possible version
#include<cstdio>
#include<cctype>
#include<cmath>
#include<algorithm>
#include<cstdlib>
#include<ctime>
#define rg register
namespace fast_IO
{
const int IN_LEN=1000000,OUT_LEN=1000000;
char ibuf[IN_LEN],obuf[OUT_LEN],*ih=ibuf+IN_LEN,*oh=obuf,*lastin=ibuf+IN_LEN,*lastout=obuf+OUT_LEN-1;
inline char getchar_(){return (ih==lastin)&&(lastin=(ih=ibuf)+fread(ibuf,1,IN_LEN,stdin),ih==lastin)?EOF:*ih++;}
inline void putchar_(const char x){if(oh==lastout)fwrite(obuf,1,oh-obuf,stdout),oh=obuf;*oh++=x;}
inline void flush(){fwrite(obuf,1,oh-obuf,stdout);}
}
using namespace fast_IO;
#define getchar() getchar_()
#define putchar(x) putchar_((x))
typedef long long ll;
template<typename T>inline T min(const T x,const T y){return x<y?x:y;}
template<typename T>inline T max(const T x,const T y){return x>y?x:y;}
template<typename T>inline T abs(const T x){return x>0?x:-x;}
template<typename T>inline T gcd(const T a,const T b){if(!b)return a;return gcd(b,a%b);}
template<typename T>inline void swap(T&a,T&b){const T c=a;a=b;b=c;}
template<typename T>inline void read(T&x)
{
char cu=getchar();bool fla=0;x=0;
while(!isdigit(cu)){if(cu=='-')fla=1;cu=getchar();}
while(isdigit(cu))x=x*10+cu-'0',cu=getchar();
if(fla)x=-x;
}
template<typename T>inline void printe(const T x)
{
if(x>=10)printe(x/10);
putchar(x%10+'0');
}
template<typename T>inline void print(const T x)
{
if(x<0)putchar('-'),printe(-x);
else printe(x);
}
const int maxn=500001;
struct Point
{
int x,key,lson,rson,size;
}Q[maxn];int usd,root;
int newnode(const int v)
{
usd++;
Q[usd]=(Point){v,rand(),0,0,1};
return usd;
}
void update(const int x)
{
Q[x].size=Q[Q[x].lson].size+Q[Q[x].rson].size+1;
}
std::pair<int,int> split(const int x,const int v)
{
std::pair<int,int>res(0,0);
if(!x)return res;
if(Q[x].x<=v)
{
res=split(Q[x].rson,v);
Q[x].rson=res.first;
res.first=x;
}
else
{
res=split(Q[x].lson,v);
Q[x].lson=res.second;
res.second=x;
}
update(x);
return res;
}
int merge(int x,int y)
{
if(!x)return y;
if(!y)return x;
if(Q[x].key<Q[y].key)
{
Q[x].rson=merge(Q[x].rson,y);
update(x);
return x;
}
else
{
Q[y].lson=merge(x,Q[y].lson);
update(y);
return y;
}
}
void insert(const int x)
{
std::pair<int,int> R=split(root,x);
root=merge(merge(R.first,newnode(x)),R.second);
}
void del(const int x)
{
std::pair<int,int> R=split(root,x),T=split(R.first,x-1);
if(T.second==0)return;
root=merge(merge(T.first,merge(Q[T.second].lson,Q[T.second].rson)),R.second);
}
int rerank(int v)
{
int x=root,ans=0;
while(x)
{
if(Q[x].x<v)ans+=Q[Q[x].lson].size+1,x=Q[x].rson;
else x=Q[x].lson;
}
return ans+1;
}
int atrank(int rk)
{
int x=root;
while(1)
{
const int all=Q[Q[x].lson].size+1;
if(all==rk)return Q[x].x;
if(all<rk)x=Q[x].rson,rk-=all;
else x=Q[x].lson;
}
}
int prec(int v)
{
int x=root,ans=0;
while(x)
{
if(Q[x].x<v)ans=Q[x].x,x=Q[x].rson;
else x=Q[x].lson;
}
return ans;
}
int succ(int v)
{
int x=root,ans=0;
while(x)
{
if(Q[x].x>v)ans=Q[x].x,x=Q[x].lson;
else x=Q[x].rson;
}
return ans;
}
int n;
int main()
{
read(n),srand(time(0));
for(rg int i=1;i<=n;i++)
{
int opt,x;read(opt),read(x);
if(opt==1)insert(x);
else if(opt==2)del(x);
else if(opt==3)print(rerank(x)),putchar('\n');
else if(opt==4)print(atrank(x)),putchar('\n');
else if(opt==5)print(prec(x)),putchar('\n');
else print(succ(x)),putchar('\n');
}
return flush(),0;
}
題解
我們發現這裏是要維護序列,那麼我們需要把split魔改一下,改成“把第1-x個數放到左子樹,其餘放到右子樹”
接下來我們考慮操作
首先我們發現初始在treap中的那些數等價於一次insert操作,所以在此不討論
接下來,講述每個操作的做法
- 插入
這個操作按一般平衡樹操作就可以了
版本1(代碼已刪除):
我的代碼中把log變成了sort的,rand值直接從上到下填即可
版本2(本文代碼):
構造笛卡爾樹,複雜度變成次數log加上數量了,減小常數
- 刪除
直接旋出來那棵樹,不過值得注意的是,需要對節點進行回收,否則空間複雜度受不了 - 修改
打個子樹修改標記即可 - 翻轉
打個子樹修改標記即可 - 求和
記錄子樹權值和即可 - 求和最大的子列
記錄子列中最大的子段,最大前綴,最大後綴
值得注意的是,由於題目所述的區間長度必須大於0,所以要注意負權點的處理
這題看起來非常的簡單,其實在具體實現過程中要注意各個標記之間的影響,需要實現好
另外,注意翻轉標記下傳是翻轉狀態,不然很容易調不出來
code
#include<cstdio>
#include<cctype>
#include<cmath>
#include<algorithm>
#include<cstdlib>
#include<ctime>
#define rg register
namespace fast_IO
{
const int IN_LEN=1000000,OUT_LEN=1000000;
char ibuf[IN_LEN],obuf[OUT_LEN],*ih=ibuf+IN_LEN,*oh=obuf,*lastin=ibuf+IN_LEN,*lastout=obuf+OUT_LEN-1;
inline char getchar_(){return (ih==lastin)&&(lastin=(ih=ibuf)+fread(ibuf,1,IN_LEN,stdin),ih==lastin)?EOF:*ih++;}
inline void putchar_(const char x){if(oh==lastout)fwrite(obuf,1,oh-obuf,stdout),oh=obuf;*oh++=x;}
inline void flush(){fwrite(obuf,1,oh-obuf,stdout);}
}
using namespace fast_IO;
#define getchar() getchar_()
#define putchar(x) putchar_((x))
typedef long long ll;
template<typename T>inline T min(const T x,const T y){return x<y?x:y;}
template<typename T>inline T max(const T x,const T y){return x>y?x:y;}
template <typename T> inline void mind(T&a,const T b){a=a<b?a:b;}
template <typename T> inline void maxd(T&a,const T b){a=a>b?a:b;}
template<typename T>inline T abs(const T x){return x>0?x:-x;}
template<typename T>inline T gcd(const T a,const T b){if(!b)return a;return gcd(b,a%b);}
template<typename T>inline void swap(T&a,T&b){const T c=a;a=b;b=c;}
template<typename T>inline void read(T&x)
{
char cu=getchar();bool fla=0;x=0;
while(!isdigit(cu)){if(cu=='-')fla=1;cu=getchar();}
while(isdigit(cu))x=x*10+cu-'0',cu=getchar();
if(fla)x=-x;
}
template<typename T>inline void printe(const T x)
{
if(x>=10)printe(x/10);
putchar(x%10+'0');
}
template<typename T>inline void print(const T x)
{
if(x<0)putchar('-'),printe(-x);
else printe(x);
}
const int maxn=500001;
struct Point//平衡樹節點
{
int x,key,lson,rson,size;
int zuoda,youda,zhong,he;
bool flag;int change;
}Q[maxn];int root;
void blbl(const int x)
{
if(!x)return;
printf("%d\n",Q[x].x);
blbl(Q[x].lson);
blbl(Q[x].rson);
}
int rubbish[maxn],top;
int newnode(const int R)//新建節點 傳入rand
{
int usd=rubbish[top--];
Q[usd]=(Point){0,R,0,0,1,0,0,0,0,0,-1001};
return usd;
}
void change(Point&hos,const int x)//改變單點的x
{
hos.x=x;
maxd(hos.zuoda,x),maxd(hos.youda,x);
hos.zhong=x;
hos.he+=x;
}
void deal_flag(Point&hos)//解決翻轉
{
swap(hos.lson,hos.rson);
swap(hos.zuoda,hos.youda);
hos.flag^=1;
}
void deal_change(Point&hos,const int x)//解決賦值
{
hos.x=x;
if(x>0)hos.zuoda=hos.youda=hos.zhong=hos.he=hos.size*x;
else hos.zuoda=hos.youda=0,hos.zhong=x,hos.he=hos.size*x;
hos.change=x;
}
void down(Point&hos)//下傳標記
{
int w=hos.lson;
if(w)
{
if(hos.flag)deal_flag(Q[w]);
if(hos.change!=-1001)deal_change(Q[w],hos.change);
}
w=hos.rson;
if(w)
{
if(hos.flag)deal_flag(Q[w]);
if(hos.change!=-1001)deal_change(Q[w],hos.change);
}
hos.flag=0,hos.change=-1001;
}
void update(Point&hos)//更新
{
Point&L=Q[hos.lson],&R=Q[hos.rson];
hos.size=L.size+R.size+1;
hos.he=L.he+R.he+hos.x;
hos.zuoda=max(L.zuoda,L.he+hos.x+R.zuoda);
hos.youda=max(R.youda,R.he+hos.x+L.youda);
hos.zhong=max(max(L.zhong,R.zhong),L.youda+hos.x+R.zuoda);
}
std::pair<int,int> split(const int x,const int kth)
{
std::pair<int,int>res(0,0);
if(!x)return res;
down(Q[x]);
if(Q[Q[x].lson].size+1==kth)
{
res.first=x,res.second=Q[x].rson;
Q[x].rson=0;
}
else if(Q[Q[x].lson].size+1<kth)
{
res=split(Q[x].rson,kth-(Q[Q[x].lson].size+1));
Q[x].rson=res.first;
res.first=x;
}
else
{
res=split(Q[x].lson,kth);
Q[x].lson=res.second;
res.second=x;
}
update(Q[x]);
return res;
}
int merge(int x,int y)
{
if(!x)return y;
if(!y)return x;
down(Q[x]),down(Q[y]);
if(Q[x].key<Q[y].key)
{
Q[x].rson=merge(Q[x].rson,y);
update(Q[x]);
return x;
}
else
{
Q[y].lson=merge(x,Q[y].lson);
update(Q[y]);
return y;
}
}
int m,P[maxn],Ptop;
char opt[11];int opttop;
int STA[maxn],Stop,ID[maxn];
int maker()
{
int rando=rand();
STA[Stop=1]=newnode(rando),ID[Stop]=P[1];
for(rg int i=2;i<=Ptop;i++)
{
int ME=newnode(rando);
rando=rand();
bool fla=0;
while(Stop>1&&rando<Q[STA[Stop]].key)Q[STA[Stop-1]].rson=STA[Stop],change(Q[STA[Stop]],ID[Stop]),update(Q[STA[Stop]]),Stop--,fla=1;
if(rando<Q[STA[Stop]].key)
{
change(Q[STA[Stop]],ID[Stop]),update(Q[STA[Stop]]);
Q[ME].lson=STA[Stop];
STA[Stop]=ME,ID[Stop]=P[i];
}
else
{
if(fla)
{
Q[STA[Stop]].rson=0;
Q[ME].lson=STA[Stop+1];
}
STA[++Stop]=ME,ID[Stop]=P[i];
}
}
while(Stop>1)Q[STA[Stop-1]].rson=STA[Stop],change(Q[STA[Stop]],ID[Stop]),update(Q[STA[Stop]]),Stop--;
change(Q[STA[Stop]],ID[Stop]),update(Q[STA[Stop]]);
return STA[1];
}
void insert(const int x)
{
std::pair<int,int>T=split(root,x);
root=merge(merge(T.first,maker()),T.second);
}
void rerubbish(const int x)
{
if(!x)return;
rubbish[++top]=x;
rerubbish(Q[x].lson),rerubbish(Q[x].rson);
}
void del(const int x)
{
std::pair<int,int>T=split(root,x),S=split(T.second,Ptop);
rerubbish(S.first);
root=merge(T.first,S.second);
}
int samv;
void sam(const int x)
{
std::pair<int,int>T=split(root,x),S=split(T.second,Ptop);
deal_change(Q[S.first],samv);
root=merge(merge(T.first,S.first),S.second);
}
void reverse(const int x)
{
std::pair<int,int>T=split(root,x),S=split(T.second,Ptop);
deal_flag(Q[S.first]);
root=merge(merge(T.first,S.first),S.second);
}
int sum(const int x)//complete
{
std::pair<int,int>T=split(root,x),S=split(T.second,Ptop);
int res=Q[S.first].he;
root=merge(merge(T.first,S.first),S.second);
return res;
}
int bigv(const int x)//compete
{
return Q[root].zhong;
}
int main()
{
Q[0].zhong=-(0x7f7f7f7f);
for(rg int i=1;i<=500000;i++)rubbish[++top]=i;
read(Ptop),read(m),srand(time(0));
for(rg int i=1;i<=Ptop;i++)read(P[i]);
insert(0);
for(rg int i=1;i<=m;i++)
{
int x;
opt[opttop=1]=getchar();
while(opt[1]<'A'||opt[1]>'Z')opt[1]=getchar();
while(opt[opttop]!=' '&&opt[opttop]!='\n'&&opt[opttop]!=EOF)opt[++opttop]=getchar();
if(opt[1]=='I')
{
read(x),read(Ptop);
for(rg int i=1;i<=Ptop;i++)read(P[i]);
insert(x);
}
else if(opt[1]=='D')
{
read(x),read(Ptop);x--;
del(x);
}
else if(opt[3]=='K')
{
read(x),read(Ptop),read(samv);x--;
sam(x);
}
else if(opt[1]=='R')
{
read(x),read(Ptop);x--;
reverse(x);
}
else if(opt[1]=='G')
{
read(x),read(Ptop);x--;
print(sum(x)),putchar('\n');
}
else print(bigv(root)),putchar('\n');
}
return flush(),0;
}
總結
一道較爲繁瑣的數據結構,調了有一會兒,馬上就要csp了,還得加強代碼熟練度