1057. Stack (30)

时间限制

150 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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

#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;//下标1位置为栈底
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;//统计小于第k小的数之前的个数
  int index;
  for (int i = 0; i < blockrange; i++)
  {
    sum += block[i];//分块法
    if (sum >=k)
    {
      index = i;
      sum -= block[i];
      break;
    }
  }
  int  start = (index)*blockrange;//这里注意块号是从0开始的,并且注意每个块的管辖范围
  for (int i = start; i < start + blockrange; i++)
  {//这里注意记录每个数出现的次数的table的下标是从1开始的
    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;
}
/*
分块法：先划分sqrt（maxn）（向上取整）个块，然后用hash表统计每个输入的数的个数
并用块表统计每个块内数字出现的总个数，注意块号从0开始，管辖范围也是从k*blocksize
开始，k=0,1,2,3...
*/