HDU - 4006 The kth great number（优先队列）

最新推荐文章于 2019-03-28 10:07:13 发布

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本文介绍了一个使用优先队列数据结构解决游戏《数字游戏》中数值查询问题的方法。游戏中，小明可以写下一个数字或询问小宝第k大的数字。为帮助小宝快速响应，文章详细阐述了如何通过维护一个大小为k的优先队列来实时更新并提供所需数值。此方法确保了即使面对大量数字输入，也能高效准确地返回第k大的数字。

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Xiao Ming and Xiao Bao are playing a simple Numbers game. In a round Xiao Ming can choose to write down a number, or ask Xiao Bao what the kth great number is. Because the number written by Xiao Ming is too much, Xiao Bao is feeling giddy. Now, try to help Xiao Bao.

Input

There are several test cases. For each test case, the first line of input contains two positive integer n, k. Then n lines follow. If Xiao Ming choose to write down a number, there will be an " I" followed by a number that Xiao Ming will write down. If Xiao Ming choose to ask Xiao Bao, there will be a "Q", then you need to output the kth great number.

Output

The output consists of one integer representing the largest number of islands that all lie on one line.

Sample Input

8 3
I 1
I 2
I 3
Q
I 5
Q
I 4
Q

Sample Output

1
2
3

Hint

Xiao  Ming  won't  ask  Xiao  Bao  the  kth  great  number  when  the  number  of  the  written number is smaller than k. (1=<k<=n<=1000000).

/*
本题是一个优先队列的题目，要求第k个大的数，所以保持队列里面
有k个元素 
*/
#include <iostream>
#include <queue>
#include <string.h>
using namespace std;
struct in
{
	int num;
}s;
bool operator < (const in &a,const in &b) //运算符重载 
{
	return a.num>b.num;
}
int main()
{
	priority_queue<in> q;
	int n,k,i,j,sum;
	char a[5];
	while(cin>>n>>k)
	{
		sum=0;
		while(n--)
		{
			cin>>a;
			if(strcmp(a,"I")==0)
			{
				if(sum<k)
				{
					cin>>s.num;
					q.push(s);
					sum++;
				}
				else
				{
					cin>>s.num;
					if(s.num>q.top().num)
					{
						q.pop();
						q.push(s);	
					}	
				} 
			}
			else
			{
				s=q.top();
				cout<<s.num<<endl;
			}
		}
		while(!q.empty())//初始化，十分重要 
		{
			q.pop();
		}
	}
	return 0;
}