HDU 4006 The kth great number(优先队列)

最新推荐文章于 2020-03-30 22:24:05 发布

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本文深入解析了如何使用优先队列解决寻找第K大的数的问题，并通过实例代码展示了具体实现过程。

第K大的数，我也是醉了，刚开始不能理解第k大的数，一直认为1,2,3这几个三大的数是3啊。真是愚蠢之极啊！！！3最大，3是第一大。

然后用一般方法会超时，所以用priority_queue<int, vector<int>, greater<int> > pq;优先队列，更改了排序为逆序用greater<int>，默认正序排序less<int>

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).

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <stack>
#include <queue>
#include <algorithm>
#include <set>
using namespace std;
#define LOCAL
#define INF 0x3f3f3f3f
#define MAX_N 10000

priority_queue<int, vector<int>, greater<int> > pq;



int main()
{
	#ifdef LOCAL
		freopen("b:\\data.in.txt", "r", stdin);
	#endif

    int n, k, in;
    char c;

    while(~scanf("%d%d", &n, &k))
    {
        while(!pq.empty()) pq.pop();
        for(int i = 0; i < k; i++) {
            cin >> c;
            scanf("%d", &in);
            pq.push(in);
        }

        int cnt = n - k;
        while(cnt--) {
            int kth = pq.top();
//            cout << pq.top() << endl;
            cin >> c;
            if(c == 'I') {
                scanf("%d", &in);
                if(in > kth) {
                    pq.pop();
                    pq.push(in);
                }
            } else {
                printf("%d\n", kth);
            }
        }
    }
    return 0;
}