PAT：Stack (30)（树状数组+二分查找）

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.

 

输入描述:

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.

 

输出描述:

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.

 

输入例子:

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

 

输出例子:

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

【解析】

最近老遇到树状数组，不得不说很有用了吧。

题意：就是栈的操作，push就推入栈中不用输出，pop就弹出栈顶元素并输出，还有一种peekmedian操作，找到第（n+1）/2【题目是说偶数n/2,奇数(n+1)/2但是其实取整一下都是(n+1)/2不难理解吧】小的数并输出。

#include 
using namespace std;
const int maxn = 100010;
using namespace std;
stack s;//栈只用来操作pop和push
int c[maxn];//c才是用来peekmedian的
int lowbit(int x)
{
	return x & (-x);
}
void update(int x, int v)
{
	for (int i = x; i < maxn; i += lowbit(i))
		c[i] += v;
}
int getsum(int x) 
{
	int sum = 0;
	for (int i = x; i >= 1; i -= lowbit(i))
		sum += c[i];
	return sum;
}
void PeekMedian() 
{
	int left = 1, right = maxn, mid, k = (s.size() + 1) / 2;
	while (left < right) 
	{
		mid = (left + right) / 2;
		if (getsum(mid) >= k)
			right = mid;
		else
			left = mid + 1;
	}
	printf("%d\n", left);
}
int main() 
{
	int n, temp;
	scanf("%d", &n);
	char str[20];
	while(n--)
	{
		scanf("%s", str);
		if (str[1] == 'u') 
		{
			scanf("%d", &temp);
			s.push(temp);
			update(temp, 1);//push就是+1
		}
		else if (str[1] == 'o')
		{
			if (!s.empty()) 
			{
				update(s.top(), -1);//删除就是-1
				printf("%d\n", s.top());
				s.pop();
			}
			else 
				printf("Invalid\n");
		}
		else 
		{
			if (!s.empty())
				PeekMedian();
			else
				printf("Invalid\n");
		}
	}
	return 0;
}