1057 Stack

实现堆栈与中位数查询

最新推荐文章于 2023-09-21 11:15:51 发布

原创最新推荐文章于 2023-09-21 11:15:51 发布 · 276 阅读

0 ·

CC 4.0 BY-SA版权

pat 同时被 2 个专栏收录

71 篇文章

订阅专栏

分块

2 篇文章

订阅专栏

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 N is 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 Invalid instead.

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

实在想不出来，太菜了自己。没想到居然还有分块的算法，实在。。。。。

解析：

首先，它会将一个序列分成 √n 块，除了最后一块，其他每一块都严格的只有 √n 个元素，然后用一个block数组记录每一个元素属于哪一块，时间复杂度为√n

#include<set>
#include<map>
#include<list>
#include<queue>
#include<deque>
#include<cmath>
#include<stack>
#include<cstdio>
#include<string>
#include<bitset>
#include<vector>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#include<bits/stdc++.h>
using namespace std;

#define e exp(1)
#define p acos(-1)
#define mod 1000000007
#define inf 0x3f3f3f3f
#define ll long long
#define ull unsigned long long
#define mem(a,b) memset(a,b,sizeof(a))
int gcd(int a,int b) {
	return b?gcd(b,a%b):a;
}

const int maxn=1e5+5;
int sqrN=361;
int block[361];
int table[maxn];
stack<int> st;

void Push(int x)
{
	st.push(x);
	block[x/sqrN]++;
	table[x]++;
}
void Pop()
{
	int t=st.top();
	st.pop();
	block[t/sqrN]--;
	table[t]--;
	printf("%d\n",t);
}
void PeekMedian(int k)
{
	int sum=0,index=0;
	while(sum+block[index]<k)
	{
		sum+=block[index++];
	}
	int num=index*sqrN;
	while(sum+table[num]<k)
	{
		sum+=table[num++];
	}
	printf("%d\n",num);
}
int main()
{
	char s[25];
	int T,x,k;scanf("%d",&T);
	while(T--)
	{
		scanf("%s",s);
		if(strcmp(s,"Push")==0)
		{
			scanf("%d",&x);
			Push(x);
		}
		else if(strcmp(s,"Pop")==0)
		{
			if(st.empty())puts("Invalid");
			else Pop();
		}
		else
		{
			if(st.empty())puts("Invalid");
			else 
			{
				int n=st.size();
				if(n%2)PeekMedian((n+1)/2);
				else PeekMedian(n/2);
			}
		}
	}
	return 0;
}