The kth great number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65768/65768 K (Java/Others)

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

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题意：n个操作，‘I’表示插入值，Q则问你第k大数是多少。

题解：

优先队列 priority_queue<int, vector<int>, greater<int> >的使用，每插入一个数都保存前k大的数，队首就为第k大的数。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<string>
#include<algorithm>
#include<cstdlib>
#include<set>
#include<queue>
#include<stack>
#include<vector>
#include<map>

#define N 100010
#define Mod 10000007
#define lson l,mid,idx<<1
#define rson mid+1,r,idx<<1|1
#define lc idx<<1
#define rc idx<<1|1
typedef long long ll;

const int INF = 1000010;

using namespace std;

int main()
{
    int n, k;
    while ( cin >> n >> k )
    {
        char c;
        int s;
        priority_queue<int, vector<int>, greater<int> >que;///从大到小排序，队首尾最小值
        while ( n-- )
        {
            scanf ( "%c%*c", &c );
            if ( c == 'Q' )
                printf ( "%d\n", que.top() );
            else
            {
                scanf ( "%d", &s );
                que.push ( s );
                if ( que.size() > k )
                    que.pop();
                getchar();
            }
        }
    }
    return 0;
}