优先队列 The kth great number

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).

这题就是不停的插入数字然后不停的排序，直接排肯定超时，所以我们用优先队列，就弄一个大小为k的优先队列，如果新插入的数字比最小的那个大的话就放进去，否则就不变，如果遇到Q，就输出最小的那个数字就行。

#include <stdio.h> #include<queue> #include<vector> using namespace std; priority_queue<int,vector<int>,greater<int> >q; int main() { int n,k; while(~scanf("%d%d",&n,&k)) { while(!q.empty()) q.pop(); int i; for(i=0;i<n;i++) { char a; scanf("%c",&a); scanf("%c",&a); if(a=='I') { int num; scanf("%d",&num); if(q.size()<k) q.push(num); else { if(q.top()<num) { q.pop(); q.push(num); } } } else { printf("%d\n",q.top()); } } } return 0; }