zoukankan      html  css  js  c++  java
  • hdu KiKi's K-Number 主席树

    KiKi's K-Number

    Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)


    Problem Description
    For the k-th number, we all should be very familiar with it. Of course,to kiki it is also simple. Now Kiki meets a very similar problem, kiki wants to design a container, the container is to support the three operations.

    Push: Push a given element e to container

    Pop: Pop element of a given e from container

    Query: Given two elements a and k, query the kth larger number which greater than a in container;

    Although Kiki is very intelligent, she can not think of how to do it, can you help her to solve this problem?
     
    Input
    Input some groups of test data ,each test data the first number is an integer m (1 <= m <100000), means that the number of operation to do. The next m lines, each line will be an integer p at the beginning, p which has three values:
    If p is 0, then there will be an integer e (0 <e <100000), means press element e into Container.

    If p is 1, then there will be an integer e (0 <e <100000), indicated that delete the element e from the container  

    If p is 2, then there will be two integers a and k (0 <a <100000, 0 <k <10000),means the inquiries, the element is greater than a, and the k-th larger number.
     
    Output
    For each deletion, if you want to delete the element which does not exist, the output "No Elment!". For each query, output the suitable answers in line .if the number does not exist, the output "Not Find!".
     
    Sample Input
    5 0 5 1 2 0 6 2 3 2 2 8 1 7 0 2 0 2 0 4 2 1 1 2 1 2 2 1 3 2 1 4
     
    Sample Output
    No Elment! 6 Not Find! 2 2 4 Not Find!
     
    Source
    #pragma comment(linker, "/STACK:1024000000,1024000000")
    #include<iostream>
    #include<cstdio>
    #include<cmath>
    #include<string>
    #include<queue>
    #include<algorithm>
    #include<stack>
    #include<cstring>
    #include<vector>
    #include<list>
    #include<set>
    #include<map>
    using namespace std;
    #define ll long long
    #define pi (4*atan(1.0))
    #define eps 1e-14
    #define bug(x)  cout<<"bug"<<x<<endl;
    const int N=1e5+10,M=1e6+10,inf=2e9;
    const ll INF=1e18+10,mod=2147493647;
    struct Chairmantree
    {
        int rt[N*20],ls[N*20],rs[N*20],sum[N*20];
        int tot;
        void init()
        {
            tot=0;
        }
        void build(int l,int r,int &pos)
        {
            pos=++tot;
            sum[pos]=0;
            if(l==r)return;
            int mid=(l+r)>>1;
            build(l,mid,ls[pos]);
            build(mid+1,r,rs[pos]);
        }
        void update(int p,int c,int pre,int l,int r,int &pos)
        {
            pos=++tot;
            ls[pos]=ls[pre];
            rs[pos]=rs[pre];
            sum[pos]=sum[pre]+c;
            if(l==r)return;
            int mid=(l+r)>>1;
            if(p<=mid)
                update(p,c,ls[pre],l,mid,ls[pos]);
            else
                update(p,c,rs[pre],mid+1,r,rs[pos]);
        }
        int rank(int s,int e,int L,int R,int l,int r)
        {
            if(L<=l&&r<=R)return sum[e]-sum[s];
            int mid=(l+r)>>1;
            int ans=0;
            if(L<=mid)
                ans+=rank(ls[s],ls[e],L,R,l,mid);
            if(R>mid)
                ans+=rank(rs[s],rs[e],L,R,mid+1,r);
            return ans;
        }
        int query(int L,int R,int l,int r,int k)
        {
            if(l==r)return l;
            int mid=(l+r)>>1;
            int x=sum[ls[R]]-sum[ls[L]];
            if(k<=x) return query(ls[L],ls[R],l,mid,k);
            else return query(rs[L],rs[R],mid+1,r,k-x);
        }
    };
    Chairmantree tree;
    int main()
    {
        int n,le=1e5+10;
        while(~scanf("%d",&n))
        {
            tree.init();
            tree.build(1,le,tree.rt[0]);
            for(int i=1;i<=n;i++)
            {
                int x;
                scanf("%d",&x);
                if(x==0)
                {
                    int z;
                    scanf("%d",&z);
                    tree.update(z,1,tree.rt[i-1],1,le,tree.rt[i]);
                }
                else if(x==1)
                {
                    int z;
                    scanf("%d",&z);
                    if(tree.rank(tree.rt[0],tree.rt[i-1],z,z,1,le))
                    {
                        tree.update(z,-1,tree.rt[i-1],1,le,tree.rt[i]);
                    }
                    else
                    {
                        tree.update(z,0,tree.rt[i-1],1,le,tree.rt[i]);
                        printf("No Elment!
    ");
                    }
                }
                else
                {
                    tree.update(1,0,tree.rt[i-1],1,le,tree.rt[i]);
                    int a,k;
                    scanf("%d%d",&a,&k);
                    int v=tree.rank(tree.rt[0],tree.rt[i],1,a,1,le);
                    k+=v;
                    int q=tree.rank(tree.rt[0],tree.rt[i],1,le,1,le);
                    //cout<<"xxxx "<<v<<" "<<k<<" "<<q<<endl;
                    if(k>q)
                        printf("Not Find!
    ");
                    else
                        printf("%d
    ",tree.query(tree.rt[0],tree.rt[i],1,le,k));
                }
                //for(int j=1;j<=9;j++)
                    //cout<<"xxx "<<j<<" "<<tree.rank(tree.rt[0],tree.rt[i],j,j,1,le)<<endl;
            }
        }
        return 0;
    }
  • 相关阅读:
    java如何编写多线程
    Java调用dll动态库
    HashMap源码解析
    LinkedList源码解析
    ArrayList源码解析
    springboot配置cxf
    java生成二维码
    原生js--跨域消息传递
    原生js--应用程序存储和离线web应用
    原生js--userData
  • 原文地址:https://www.cnblogs.com/jhz033/p/6502538.html
Copyright © 2011-2022 走看看