P3369 [template] general balanced tree 01Trie tree

P3369 [template] general balanced tree

Title Description

You need to write a data structure (refer to the subject of the title), to maintain some of the numbers, which need to provide the following:

1. Insert x number x
2. Delete the X- the X-number (the same number if more than one, because only a delete)
3. Query x rank number x (rank number is defined as the ratio of the number of the current number of small + 1 + if the same number of a plurality of 1, because the minimum output rank)
4. Queries ranked x number x
5. Seeking X X precursor (precursor defined as less than X X, and the maximum number)
6. Seeking X X successor (defined as greater than the subsequent X X, and the smallest number)

Input Format

The first line n the number n, the operation, the following n n lines each have two numbers opt O P T and X X, opt O P T represents the serial number of the operation (  . 1 \ Leq opt \ Leq. 6 . 1 ≤ O P t ≤ 6)

Output Format

For operation 3,4,5,6 3 , 4 , 5 , 6 a number of outputs per row, indicates that the corresponding answer

Sample input and output

Input # 1copy

10
1 106465
4 1
1 317721
1 460929
1 644985
1 84185
1 89851
6 81968
1 492737
5 493598

Output # 1copy

106465
84185
492737

Description / Tips

Constraints of time: 1000ms, 128M

1.n data ranges:  n-\ 100000 Leq n- ≤ . 1 0 0 0 0 0

2. Each of the number of data ranges: [- 7 ^ {10}, {10 ^ 7}] [ - . 1 0 7 , . 1 0 7 ]

Source: Tyvj1728 original name: Plain balanced tree

In Acknowledgments

#include <bits / STDC ++ H.>
 #define INF 1E7
 the using  namespace STD;
 const  int MAXN = 100005 , MAXM MAXN * = 20 is , N = 24 ; // 2 ^ 20 is 
int CH [MAXM] [ 2 ], V [MAXM ];
 int CNT = . 1 ;
 void the Add ( int POS, int QX) 
{ 
    POS + = INF; // prevent negative 
    int P = . 1 ; // the vacant node number 0 
    for ( int I = N; I> = 0 ; i--) 
    { 
        BOOL B = (POS >> I) & . 1 ; // judgment is not. 1 
        IF (! CH [P] [B]) 
            CH [P] [B] = CNT ++; // dynamic prescription 
        v [ P] + = QX; 
        P = CH [P] [B]; 
    } 
    V [P] + = QX; // leaf nodes are not modified 
}
 int the Sum ( int POS) 
{ 
    POS + = + INF . 1 ; // + 1 is less than or equal the 
    int RES = 0 , P = 1 ;
     for ( int I = N; I> = 0;i--)
    {
        bool b=(pos>>i)&1;
        if(b)// right
            res+=v[ch[p][0]];
        p=ch[p][b];
        if(!p)
            break;
    }
    return res;
}
int Get(int rank)
{
    int p=1,res=0;
    for(int i=N;i>=0;i--)
    {
        if(v[ch[p][0]]>=rank)
            p=ch[p][0];
        else
            rank-=v[ch[p][0]],p=ch[p][1],res|=1<<i;
    }
    return res-inf;
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        int opt,x;
        scanf("%d%d",&opt,&x);
        if(opt==1)
        {
            Add(x,1);
        }
        if(opt==2)
        {
            Add(x,-1);
        }
        if(opt==3)
        {
            printf("%d\n",Sum(x-1)+1);
        }
        if(opt==4)
        {
            printf("%d\n",Get(x));
        }
        if(opt==5)
        {
            int tmp=Sum(x-1)+1;//rank x 
            printf("%d\n",Get(tmp-1));
        }
        if(opt==6)
        {
            int tmp=Sum(x)+1;//rank x+1
            printf("%d\n",Get(tmp)); 
        }
    }
    
    
    return 0;
}

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