pat 甲级 1057 Stack(30) (树状数组+二分)

最新推荐文章于 2020-10-05 23:39:21 发布

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本文介绍了一个特殊的栈数据结构实现，除了基本的push和pop操作外，还支持peekMedian操作，用于返回栈中所有元素的中位数。通过使用线段树和二分查找，文章详细阐述了如何高效地进行中位数查询，即使在频繁的插入和删除操作下也能保持良好的性能。

1057 Stack （30 分）

Stack is one of the most fundamental data structures, which is based on the principle of Last In First Out (LIFO). The basic operations include Push (inserting an element onto the top position) and Pop (deleting the top element). Now you are supposed to implement a stack with an extra operation: PeekMedian -- return the median value of all the elements in the stack. With N elements, the median value is defined to be the (N/2)-th smallest element if N is even, or ((N+1)/2)-th if Nis odd.

Input Specification:

Each input file contains one test case. For each case, the first line contains a positive integer N (≤105). Then N lines follow, each contains a command in one of the following 3 formats:

Push key
Pop
PeekMedian

where key is a positive integer no more than 105.

Output Specification:

For each Push command, insert key into the stack and output nothing. For each Pop or PeekMedian command, print in a line the corresponding returned value. If the command is invalid, print Invalidinstead.

Sample Input:

17
Pop
PeekMedian
Push 3
PeekMedian
Push 2
PeekMedian
Push 1
PeekMedian
Pop
Pop
Push 5
Push 4
PeekMedian
Pop
Pop
Pop
Pop

Sample Output:

Invalid
Invalid
3
2
2
1
2
4
4
5
3
Invalid

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int st[100005];
ll bit[1000005];
ll n;
void add(ll i,ll x)
{
    while(i<=n)
    {
        bit[i]+=x;
        i+=i&-i;
    }
}
ll sum(ll i)
{
    ll s=0;
    while(i>0)
    {
        s+=bit[i];
        i-=i&-i;
    }
    return s;
}
int pos=0;
int main()
{
    n=100000;
    //freopen("in.txt","r",stdin);
    int N;
    cin>>N;
    string op;
    while(N--)
    {
        cin>>op;
        if(op[1]=='o')
        {
            if(pos==0){cout<<"Invalid"<<endl;continue;}
            cout<<st[pos]<<endl;
            add(st[pos],-1);
            pos--;
        }
        else if(op[1]=='u')
        {
            int num;
            cin>>num;
            add(num,1);
            st[++pos]=num;
        }
        else if(op[1]=='e')
        {
            if(pos==0){cout<<"Invalid"<<endl;continue;}
            int x=(pos%2==0)?pos/2:(pos+1)/2;
            int lb=0;int ub=100001;
            while(ub-lb>1)
            {
                int mid=(ub+lb)/2;
                if(sum(mid)<x)lb=mid;
                else ub=mid;
            }
            cout<<ub<<endl;
        }
        else {
            cout<<"Invalid"<<endl;continue;
        }
    }
    return 0;
}