SPOJ 3273 Order statistic set

红果果的treap模板题。。。白书版treap

Order statistic set

Time Limit: 2000MS Memory Limit: Unknown 64bit IO Format: %lld & %llu

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Description

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In this problem, you have to maintain a dynamic set of numbers which support the two fundamental operations

• INSERT(S,x): if x is not in S, insert x into S
• DELETE(S,x): if x is in S, delete x from S

and the two type of queries

• K-TH(S) : return the k-th smallest element of S
• COUNT(S,x): return the number of elements of S smaller than x

Input

• Line 1: Q (1 ≤ Q ≤ 200000), the number of operations
• In the next Q lines, the first token of each line is a character I, D, K or C meaning that the corresponding operation is INSERT, DELETE, K-TH or COUNT, respectively, following by a whitespace and an integer which is the parameter for that operation.

If the parameter is a value x, it is guaranteed that 0 ≤ |x| ≤ 10⁹. If the parameter is an index k, it is guaranteed that 1 ≤ k ≤ 10⁹.

Output

For each query, print the corresponding result in a single line. In particular, for the queries K-TH, if k is larger than the number of elements in S, print the word 'invalid'.

Sample Input

Input8I -1I -1I 2C 0K 2D -1K 1K 2Output122invalid

Source

© VNOI

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#include <iostream>#include <cstdio>#include <cstring>#include <cstdlib>#include <ctime>#include <algorithm>using namespace std;const int INF=0x3f3f3f3f;struct Node{    Node * ch[2];    int r,s,v;    int cmp(int x)    {        if(x==v) return -1;        return (x<v)?0:1;    }    void maintain()    {        s=1+ch[0]->s+ch[1]->s;    }};Node* null ;void init(){    null=new Node();    null->s=0;    null->ch[0]=null->ch[1]=null;    srand(time(NULL));}void rotate(Node* &o,int d){    Node* k=o->ch[1^d];    o->ch[1^d]=k->ch[d];    k->ch[d]=o;    o->maintain();    k->maintain();    o=k;}void insert(Node* &o,int x){    if(o==null)    {        o=new Node();        o->ch[0]=o->ch[1]=null;        o->v=x;o->r=rand();o->s=1;    }    else    {        int d=o->cmp(x);        if(d==-1) return ;        insert(o->ch[d],x);        if(o->r < o->ch[d]->r) rotate(o,d^1);    }    o->maintain();}void remove(Node* &o,int x){    int d=o->cmp(x);    if(d==-1)    {        Node* u=o;        if(o->ch[0]!=null&&o->ch[1]!=null)        {            int d2=(o->ch[0]->r>o->ch[1]->r)?1:0;            rotate(o,d2);            remove(o->ch[d2],x);        }        else        {            if(o->ch[0]==null) o=o->ch[1];            else o=o->ch[0];            delete u;        }    }    else remove(o->ch[d],x);    if(o!=null) o->maintain();}int find(Node* o,int x){    while(o!=null)    {        int d=o->cmp(x);        if(d==-1) return 1;        o=o->ch[d];    }    return 0;}int Kth(Node* o,int k)///kth small{    while(o!=null)    {        int s=o->ch[0]->s;        if(k==s+1) return o->v;        else if(k<=s) o=o->ch[0];        else        {            o=o->ch[1]; k=k-s-1;        }    }    return -INF;}int Count(Node* o,int x){    int ret=0;    while(o!=null)    {        if(x>o->v)        {            ret+=o->ch[0]->s+1;            o=o->ch[1];        }        else o=o->ch[0];    }    return ret;}int main(){    int q;    scanf("%d",&q);    init();    Node* root=null;    while(q--)    {        char cmd[3];int nb;        scanf("%s%d",cmd,&nb);        if(cmd[0]=='I'&&find(root,nb)==0)        {            insert(root,nb);        }        else if(cmd[0]=='D'&&find(root,nb)==1)        {            remove(root,nb);        }        else if(cmd[0]=='K')        {            int temp=Kth(root,nb);            if(temp==-INF) puts("invalid");            else printf("%d\n",temp);        }        else if(cmd[0]=='C')        {            printf("%d\n",Count(root,nb));        }    }    return 0;}