1057. Stack (30)

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:

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.

17PopPeekMedianPush 3PeekMedianPush 2PeekMedianPush 1PeekMedianPopPopPush 5Push 4PeekMedianPopPopPopPopSample Output:
InvalidInvalid322124453Invalid
#include<cstdio>#include<iostream>#include<vector>#include<string>#include<cstring>#include<algorithm>using namespace std;const int maxn = 100000 + 10;int Stack[maxn];int top =0;//下標(biāo)1位置為棧底int N;char command[20];int blockrange = sqrt(maxn);int block[1000] = { 0 };int table[maxn] = { 0 };void push(int x){	Stack[++top] = x;}int popvalue;bool pop(){	if (top)	{		popvalue = Stack[top];		table[Stack[top]]--;		block[Stack[top] / blockrange]--;		top--;		return true;	}	return false;}int peekmedian(){	int k = (top % 2 == 0) ? top / 2 : (top + 1) / 2;	int sum = 0;//統(tǒng)計(jì)小于第k小的數(shù)之前的個(gè)數(shù)	int index;	for (int i = 0; i < blockrange; i++)	{		sum += block[i];//分塊法		if (sum >=k)		{			index = i;			sum -= block[i];			break;		}	}	int	start = (index)*blockrange;//這里注意塊號(hào)是從0開始的,并且注意每個(gè)塊的管轄范圍	for (int i = start; i < start + blockrange; i++)	{//這里注意記錄每個(gè)數(shù)出現(xiàn)的次數(shù)的table的下標(biāo)是從1開始的		sum += table[i];		if (sum >= k)		{			return i;		}	}}int main(){	scanf("%d",&N); int temp;	for (int i = 0; i < N; i++)	{		scanf("%s", command);		if (strcmp(command,"Pop")==0)		{			if (!pop())			{				printf("Invalid/n");			}			else				printf("%d/n", popvalue);		}		else if (strcmp(command, "PeekMedian") == 0)		{			if (top>0)				printf("%d/n", peekmedian());			else				printf("Invalid/n");		}		else		{			scanf("%d", &temp);			table[temp]++;			block[temp / blockrange]++;			push(temp);		}	}	return 0;}/*分塊法：先劃分sqrt（maxn）（向上取整）個(gè)塊，然后用hash表統(tǒng)計(jì)每個(gè)輸入的數(shù)的個(gè)數(shù)并用塊表統(tǒng)計(jì)每個(gè)塊內(nèi)數(shù)字出現(xiàn)的總個(gè)數(shù)，注意塊號(hào)從0開始，管轄范圍也是從k*blocksize開始，k=0,1,2,3...*/