• poj-1442 Black Box(Treap)


    题目链接:

    Black Box

    题意:

    给一个序列,m个询问,每个询问是求前x个数中的第i小是多少;

    思路:

    Treap的入门题目;Treap能实现STL的set实现不了的功能,如名次树(rank tree)rank tree 支持两种新操作,一个是找出第k小元素,一个是找出x的的rank;

    AC代码:

    #include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <algorithm>
    #include <cmath>
    //#include <bits/stdc++.h>
    #include <stack>
    #include <map>
     
    using namespace std;
     
    #define For(i,j,n) for(int i=j;i<=n;i++)
    #define mst(ss,b) memset(ss,b,sizeof(ss));
     
    typedef  long long LL;
     
    template<class T> void read(T&num) {
        char CH; bool F=false;
        for(CH=getchar();CH<'0'||CH>'9';F= CH=='-',CH=getchar());
        for(num=0;CH>='0'&&CH<='9';num=num*10+CH-'0',CH=getchar());
        F && (num=-num);
    }
    int stk[70], tp;
    template<class T> inline void print(T p) {
        if(!p) { puts("0"); return; }
        while(p) stk[++ tp] = p%10, p/=10;
        while(tp) putchar(stk[tp--] + '0');
        putchar('
    ');
    }
     
    const LL mod=1e9+7;
    const double PI=acos(-1.0);
    const int inf=1e9+10;
    const int N=1e6+10;
    const int maxn=1e3+20;
    const double eps=1e-12;
    
    int a[N];
    struct Treap
    {
        Treap *ch[2];
        int siz,key,fix;
        Treap(int key)
        {
            siz=1;
            fix=rand();
            this->key=key;
            ch[0]=ch[1]=NULL;
        }
        int cmp(int x) const
        {
            if(x==key)return -1;
            return x<key? 0:1;
        }
        void maintain()
        {
            siz=1;
            if(ch[0]!=NULL)siz+=ch[0]->siz;
            if(ch[1]!=NULL)siz+=ch[1]->siz;
        }
    };
    
    void Rotate(Treap* &o,int d)
    {
        Treap* k=o->ch[d^1];
        o->ch[d^1]=k->ch[d];
        k->ch[d]=o;
        o->maintain();
        k->maintain();
        o=k;
    }
    
    void insert(Treap* &o,int x)
    {
        if(o==NULL)o=new Treap(x);
        else 
        {
            int d=(x<o->key? 0:1);
            insert(o->ch[d],x);
            if(o->ch[d]->fix > o->fix)Rotate(o,d^1);
        }
        o->maintain();
    }
    int  getans(Treap* &o,int x)
    {
        //cout<<o->key<<"*****
    ";
        if(o==NULL||x<=0||x> o->siz)return 0;
        int s=(o->ch[0] == NULL? 0: o->ch[0]->siz);
        if(x==s+1)return o->key;
        else if(x<=s)return getans(o->ch[0],x);
        else return getans(o->ch[1],x-s-1);
    }
    void removetree(Treap* &o)
    {
        if(o->ch[0]!=NULL)removetree(o->ch[0]);
        if(o->ch[1]!=NULL)removetree(o->ch[1]);
        delete o;
        o=NULL;
    }
    
    int main()
    {
        int n,m;
        while(scanf("%d%d",&n,&m)!=EOF)
        {
            For(i,1,n)read(a[i]);
            int pre=1,x;
            Treap* root=NULL;
            For(i,1,m)
            {
                read(x);
                for(int j=pre;j<=x;j++)insert(root,a[j]);
                printf("%d
    ",getans(root,i));
                pre=x+1;
            }
            removetree(root);
        }
        return 0;
    }
    

      

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  • 原文地址:https://www.cnblogs.com/zhangchengc919/p/5796073.html
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