【HDU】3253 - The kth great number（STL，队列）

本文链接：https://blog.youkuaiyun.com/wyg1997/article/details/52026954

本文介绍了一种解决特定游戏问题的算法，通过玩家在游戏中写下数字并询问第K大的数值，利用C++中的vector和优先队列实现两种高效求解第K大数的方法。

The kth great number

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65768/65768 K (Java/Others)
Total Submission(s): 9917 Accepted Submission(s): 3947

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).

Source

The 36th ACM/ICPC Asia Regional Dalian Site —— Online Contest

用vector存数，我这里按递增顺序来存，这样就可以直接用upper_bound ( ) 函数插入数据了。

（最下面还有一种简单的方法）

代码如下：

#include <cstdio>
#include <vector>
#include <algorithm>
using namespace std;
int main()
{
	int n,k;
	char op[4];
	int num;
	int pos;
	while (~scanf ("%d %d",&n,&k))
	{
		vector<int> q;
		for (int i = 1 ; i <= n ; i++)
		{
			scanf ("%s",op);
			if (op[0] == 'I')
			{
				scanf ("%d",&num);
				if (q.empty())
				{
					q.push_back(num);
					continue;
				}
				pos = upper_bound(q.begin() , q.end() , num) - q.begin();
				q.insert(q.begin() + pos , num);
			}
			else
			{
				pos = q.end() - k - q.begin();
				printf ("%d\n",q[pos]);
			}
		}
	}
	return 0;
}

被虐了。智商高的人都不用管什么是 vector。

下面提供了另外一种做法，只保留 k 个数，从小到大排，最顶的数就是需要的数。

代码如下：

#include <cstdio>
#include <algorithm>
#include <queue>
#include <vector>
using namespace std;
struct node
{
	int v;
	bool friend operator < (node a,node b)
	{
		return a.v > b.v;
	}
}a;
int main()
{
	int n,k;
	char op[4];
	while (~scanf ("%d %d",&n,&k))
	{
		priority_queue<node> q;
		for (int i = 1 ; i <= n ; i++)
		{
			scanf ("%s",op);
			if (op[0] == 'I')
			{
				int t;
				scanf ("%d",&t);
				if (q.size() == k)
				{
					a = q.top();
					if (t > a.v)
					{
						q.pop();
						a.v = t;
						q.push(a);
					}
				}
				else
				{
					a.v = t;
					q.push(a);
				}
			}
			else
				printf ("%d\n",q.top());
		}
	}
	return 0;
}