pat a1057 栈【分块思想】

最新推荐文章于 2022-06-27 21:45:53 发布

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最新推荐文章于 2022-06-27 21:45:53 发布

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分类专栏： # ★、算法笔记 # 3、常用技巧

本文链接：https://blog.youkuaiyun.com/OpenStack_/article/details/104150680

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3、常用技巧

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本文介绍了一种使用栈数据结构实现基本操作（Push、Pop）的同时，能够高效查询当前栈内元素中位数的方法。通过将元素分布于多个预定义大小的桶中，并维护每个桶内的元素计数，可以快速定位到中位数所在的桶，进而找到确切的中位数值。此方法适用于大量数据的实时中位数查询场景。

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题目：

https://www.nowcoder.com/pat/5/submission-detail/64563013

https://pintia.cn/problem-sets/994805342720868352/problems/994805417945710592

一、问题描述

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

给出一个栈的入栈（Push）、出栈（Pop）过程，并随时通过PeekMedian命令要求输出栈中中间大小的数（Pop命令输出出栈的数）。当栈中没有元素时，Pop命令和PeekMedian命令都应该输出Invalid。

二、算法实现

1、

#include<cstdio> 
#include<cstring>
#include<stack>
using namespace std;

const int maxn=100010;
const int sqrN=316;

stack<int> st;
int block[sqrN];
int table[maxn];

void peekMedian(int K)
{
	int sum=0;
	int idx=0;  //块号 
	while(sum+block[idx]<K)
	{
		sum+=block[idx++];
	}
	int num=idx*sqrN;
	while(sum+table[num]<K)
	{
		sum+=table[num++];
	} 
	printf("%d\n",num);
}

void Push(int x)
{
	st.push(x);
	block[x/sqrN]++;
	table[x]++;
}

void Pop()
{
	int x=st.top();
	st.pop();
	block[x/sqrN]--;
	table[x]--;
	printf("%d\n",x);	
}

int main()
{
	int x,query;
	memset(block,0,sizeof(block));
	memset(table,0,sizeof(table));
	char cmd[20];
	scanf("%d",&query);
	
	for(int i=0;i<query;i++)
	{
		scanf("%s",cmd);
		if(strcmp(cmd,"Push")==0)
		{
			scanf("%d",&x);
			Push(x);
		}
		else if(strcmp(cmd,"Pop")==0)
		{
			if(st.empty()==true)
			{
				printf("Invalid\n");
			}
			else
			{
				Pop();
			}
		}
		else
		{
			if(st.empty()==true)
			{
				printf("Invalid\n");
			}
			else
			{
				int K=st.size();
				if(K%2==1) K=(K+1)/2;
				else K=K/2;
				peekMedian(K);
			}
		}		
	}
	return 0;
}

2、

#include <cstdio>
#include <stack>
#include <cstring>
using namespace std;
const int MAX = 100001;
const int MAXN = 317;
int n, block[MAXN] = {0}, table[MAX] = {0};
char order[11];
stack<int> s;
 
bool cmp(int a, int b)
{
	return a < b;
}
 
int main()
{
	scanf("%d", &n);
	while(n--)
	{
		scanf("%s", order);
		if(strcmp(order, "Pop") == 0)
		{
			if(!s.empty())
			{
				int k = s.top();
				printf("%d\n", k);
				s.pop();
				block[k/(MAXN-1)]--;
				table[k]--;
			}
			else printf("Invalid\n");
		}
		else if(strcmp(order, "PeekMedian") == 0)
		{
			if(s.empty())
			{
				printf("Invalid\n");
				continue;
			}
			int k, idx = 0, index = 0;
			if(s.size() % 2 == 0) k = s.size()/2;
			else k = (s.size()+1)/2;
			int sum = 0;			
			while(sum + block[idx] < k)
			{
				sum += block[idx++];
			}
			index = idx * (MAXN-1);
			while(sum + table[index] < k)
			{
				sum += table[index++];
			}
			printf("%d\n", index);
		}
		else
		{
			int k;
			scanf("%d", &k); 
			s.push(k);
			table[k]++;
			block[k/(MAXN-1)]++;
		}
	}
	return 0;
}