hdu40069(优先级队列)

最新推荐文章于 2021-04-01 17:21:02 发布

晓风残月xj

最新推荐文章于 2021-04-01 17:21:02 发布

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分类专栏：基本数据结构 STL 文章标签：数据结构优先级队列

本文链接：https://blog.youkuaiyun.com/xiaofengcanyuexj/article/details/9570583

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本文介绍了一种求解已写数字中第K大的数的高效算法，通过使用优先级队列实现，确保了时间复杂度从O(n*n)降低到接近O(N*logK)。该算法适用于大量数据处理场景，例如实时排名系统等。

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The kth great number

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65768/65768 K (Java/Others)
Total Submission(s): 5050 Accepted Submission(s): 2070

Problem Description

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).
 
 
本题即求所写数的第k大数，1=<k<=n<=1000000，由于n,k都比较大，不能用传统排序方法做，即便只存储k个数在处理也是不合理的，需要利用优先级队列来降低时间复杂度O(n*n)变成O(N^log)，将队列中元素从小到大排序，若队列元素个数小于k,直接插入；若队列元素等于k，则需比较k与队头元素，若比队头元素小，不插入，反之则队头元素出队列，将新元素插入，保证队头元素总是第K大。
 
#include<iostream>
#include<cstdio>
#include<queue>
#include<cstring>
using namespace std;

struct cmp
{
	bool operator()(int a,int b)
	{
		return a>b;
	}
};

int main()
{
	char cmd[5];
	int i,n,k,a;
	while(~scanf("%d%d",&n,&k))
	{
		priority_queue<int,vector<int>,cmp>qq;
		for(i=0;i<n;i++)
		{
			scanf("%s",cmd);
			if(cmd[0]=='I')
			{
				scanf("%d",&a);
				if(qq.size()<k)
					qq.push(a);
				else 
				{
					int tmp=qq.top();
					if(a<tmp)
						continue;
					qq.pop();
					qq.push(a);
				}
			//	cout<<qq.top()<<"  "<<endl;
			}
			else
			{
				printf("%d\n",qq.top());
			}
		}
	}
	return 0;
}