HDU 4006 - The kth great number(优先队列)

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本文介绍了一个基于优先队列实现的算法，用于解决在一系列动态输入的整数中找到第K大的数的问题。通过维护一个大小为K的降序优先队列，可以高效地获取当前序列中第K大的数值。

The kth great number

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65768/65768 K (Java/Others)
Total Submission(s): 10062 Accepted Submission(s): 4015

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

优先队列思路:
根据来的先后顺序维护优先队列(降序),保证队列的元素个数等于K,这样队首元素就是结果.

AC代码

#include<stdio.h>
#include<queue>
#include<vector>
using namespace std;
struct mycmp
{
    bool operator()(const int &a,const int &b)
    {
        return a>b;//表示优先队列的队尾最小
    }
};
int main()
{
    int n;
    int k;
    while(~scanf("%d%d",&n,&k))
    {
        getchar();
        priority_queue,mycmp>qdu;//建立优先队列
        while(n--)
        {
            char temp[2];
            int a;
            scanf("%s",temp);
            if(temp[0] == 'I')
            {
                scanf("%d",&a);
                qdu.push(a);//入队
                while(qdu.size()>k)
                    qdu.pop();//保证队列中元素的个数一直为K
            }
            else
                printf("%d\n",qdu.top());//输出队尾元素
        }
    }
    return 0;
}