HDU4006 The kth great number————栈和队列（优先队列）

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$k. (1=<k<=n<=1000000).$

---

题意:
给你一些数字，问这些数字中，第k大的数字是谁
第一行n和k
n代表n个操作，I表示给你数字，Q 就是让你输出给你的这些数字中第k大的是几
下边n行，有n个操作

---

思路:
这道题用数组模拟和用map都不能过，总是TLE
最后用优先队列才过，优先队列真是强

我们让每次给的数字都存到优先队列中，队列的元素值越小，优先级越高，我们只需要满足队列中有k个元素就行了

队列中始终存留着最大的k个元素

---

#include<cstdio>
#include<cstring>
#include<string>
#include<queue>
#include<iostream> 
#include<map>
#include<vector>
#include<stack>
#include<algorithm>
using namespace std;
typedef long long ll;

int main()
{
    ios::sync_with_stdio(false);
    int n,k;
    while(cin>>n>>k)
    {
        priority_queue<int,vector<int>, greater<int> > q;
        map<int ,int > m;
        char c;
        int z=0;
        for(int i=0;i<n;i++)
        {
            cin>>c;
            if(c=='I')
            {
                cin>>z;
                q.push(z);
                while(q.size()>k) q.pop();
            }
            else//询问第k大的数 
                cout<<q.top()<<endl;
        }
    }
    return 0;
}