hduoj4006 The kth great number

cloudy_happy

于 2018-07-27 13:18:50 发布

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分类专栏：优先队列

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本文介绍了一个简单的数字游戏算法，通过使用优先队列来帮助角色小宝在游戏中快速找到当前第K大的数字。该算法适用于当小明询问时，能够即时给出答案。

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The kth great number

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).
        
思路：

这道题其实很简单，使用优先队列，一直保持队中只有k个元素即可，可是遇到一个问题dev一直报错，一直没想明白（欢迎大佬解释）问题会在代码中说明；

代码：

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<string>
#include<cstring>
#include<stack>
#include<queue>
#include<vector>
#include<map>
#include<set>
#define ll long long
using namespace std;
int main()
{
   int n,k,num;
   while (~scanf("%d %d",&n,&k))
   {
       priority_queue<int,vector<int>,greater<int> > q;
       char ch[10];//这里的ch不能定义为单个字符，否则在输出队首的时候就会报错，具体原因我不知道，希望得到解答
       for (int i = 0;i < n;i ++)
       {
           scanf("%s",ch);
           if (ch[0] == 'I')
           {
               scanf("%d",&num);
               q.push(num);
               if (q.size() > k)
                   q.pop();
           }
           else
           {
               int a = q.top();
               printf("%d\n",a);
           }
       }
   }
   return 0;
}