hdoj 4006 The kth great number【优先队列】

最新推荐文章于 2018-09-18 17:14:15 发布

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最新推荐文章于 2018-09-18 17:14:15 发布

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本文介绍了一种基于优先队列的数据结构实现的算法，用于快速查找序列中第K大的数值。通过不断维护一个大小为K的优先队列，确保在任何时候都能高效地给出答案。

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The kth great number

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

Total Submission(s): 8578 Accepted Submission(s): 3378

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).

真心看累了，水一道。

题意：求一个序列里面第k大的数。

思路：用优先队列<greater>。在序列数小于k时，不会有查询。当序列数大于k时，那些第k+1、k+2...大的数（即队列前面的数）永远不会用到，可以剔除，最后队列保留k个数，查询时直接输出队列头部的数即可。

AC代码：

#include <cstdio>
#include <cstring>
#include <vector>
#include <queue>
#include <set>
using namespace std;
int main()
{
    char op[10];
    int N, k, a;
    while(scanf("%d%d", &N, &k) != EOF)
    {
        priority_queue<int,vector<int>,greater<int> >Q;
        while(N--)
        {
            scanf("%s", op);
            if(op[0] == 'I')
            {
                scanf("%d", &a);
                Q.push(a);
                if(Q.size() > k)//取掉队列前面的数 不会影响第k大数的查询
                    Q.pop();
            }
            else
                printf("%d\n", Q.top());
        }
    }
    return 0;
}