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「bzoj3324」普通平衡树 - treap | Bill Yang's Blog

路终会有尽头,但视野总能看到更远的地方。

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「bzoj3324」普通平衡树 - treap

题目大意

    您需要写一种数据结构(可参考题目标题),来维护一些数,其中需要提供以下操作:
    $1$. 插入$x$数
    $2$. 删除$x$数(若有多个相同的数,因只删除一个)
    $3$. 查询$x$数的排名(若有多个相同的数,因输出最小的排名)
    $4$. 查询排名为$x$的数
    $5$. 求$x$的前驱(前驱定义为小于$x$,且最大的数)
    $6$. 求$x$的后继(后继定义为大于$x$,且最小的数)


题目分析

来填坑啦!

之前写了splay和替罪羊树,现在来补一份treap的模板。

注意为了维持$O(n)$的内存,我使用了垃圾回收。


代码

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#include<algorithm>
#include<iostream>
#include<iomanip>
#include<cstring>
#include<cstdlib>
#include<climits>
#include<vector>
#include<cstdio>
#include<cmath>
#include<queue>
using namespace std;

inline const int Get_Int() {
int num=0,bj=1;
char x=getchar();
while(x<'0'||x>'9') {
if(x=='-')bj=-1;
x=getchar();
}
while(x>='0'&&x<='9') {
num=num*10+x-'0';
x=getchar();
}
return num*bj;
}

const int maxn=100005;

mt19937 g(rand());

struct Tree {
int child[2];
int d,val;
int size;
};

struct Treap {
Tree tree[maxn];
int size,root;
queue<int>Q;
#define d(x) tree[x].d
#define val(x) tree[x].val
#define ls(x) tree[x].child[0]
#define rs(x) tree[x].child[1]
#define root(x) tree[x].root
#define size(x) tree[x].size
void push_up(int index) {
int ls=ls(index),rs=rs(index);
size(index)=size(ls)+size(rs)+1;
}
int newnode(int val) {
int now;
if(!Q.empty()) {
now=Q.front();
Q.pop();
ls(now)=rs(now)=0;
} else now=++size;
val(now)=val;
d(now)=g()%maxn;
size(now)=1;
return now;
}
void rotate(int& index,bool side) {
int son=tree[index].child[side^1];
tree[index].child[side^1]=tree[son].child[side];
tree[son].child[side]=index;
push_up(index);
push_up(son);
index=son;
}
void insert(int& index,int val) {
if(!index) {
index=newnode(val);
return;
}
bool side=val>val(index);
int &son=tree[index].child[side];
insert(son,val);
size(index)++;
if(d(index)<d(son))rotate(index,side^1);
}
int cnt,a[maxn];
void dfs(int index) {
if(!index)return;
if(ls(index))dfs(ls(index));
a[++cnt]=val(index);
Q.push(index);
if(rs(index))dfs(rs(index));
}
int build(int Left,int Right) {
if(Left>Right)return 0;
int mid=(Left+Right)>>1,now=newnode(a[mid]);
ls(now)=build(Left,mid-1);
rs(now)=build(mid+1,Right);
push_up(now);
return now;
}
int rebuild(int index) {
cnt=0;
dfs(ls(index));
dfs(rs(index));
int pos=build(1,cnt);
return pos;
}
void remove(int val) {
int now=root,father;
bool side;
while(val(now)!=val) {
father=now;
size(now)--;
if(val<val(now))now=ls(now),side=0;
else now=rs(now),side=1;
}
int pos=rebuild(now);
if(now!=root)tree[father].child[side]=pos;
else root=pos;
}
int rank(int val) {
int now=root,ans=0;
while(now) {
if(val<=val(now))now=ls(now);
else {
ans+=size(ls(now))+1;
now=rs(now);
}
}
return ans+1;
}
int kth(int rank) {
int now=root;
while(now>0&&rank>=0) {
if(ls(now)&&size(ls(now))>=rank)now=ls(now);
else {
if(rank<=size(ls(now))+1)return now;
rank-=size(ls(now))+1;
now=rs(now);
}
}
return -1;
}
int pre(int num) {
int now=root,ans=-INT_MAX;
while(now) {
if(val(now)<num)ans=max(ans,val(now)),now=rs(now);
else now=ls(now);
}
return ans;
}
int suc(int num) {
int now=root,ans=INT_MAX;
while(now) {
if(val(now)>num)ans=min(ans,val(now)),now=ls(now);
else now=rs(now);
}
return ans;
}
} treap;

int n;

int main() {
n=Get_Int();
for(int i=1; i<=n; i++) {
int order=Get_Int();
if(order==1)treap.insert(treap.root,Get_Int());
if(order==2)treap.remove(Get_Int());
if(order==3)printf("%d\n",treap.rank(Get_Int()));
if(order==4)printf("%d\n",treap.tree[treap.kth(Get_Int())].val);
if(order==5)printf("%d\n",treap.pre(Get_Int()));
if(order==6)printf("%d\n",treap.suc(Get_Int()));
}
return 0;
}
姥爷们赏瓶冰阔落吧~