HDU 4006 The kth great number (堆实现优先队列)-优快云博客

本文介绍了一种高效算法，用于解决寻找一组数中的第K大数的问题。通过使用最小堆实现优先队列，该算法能在大量输入数据中快速找到答案，并详细展示了具体的实现过程。

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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): 4447 Accepted Submission(s): 1837

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

/*
  堆其实就相当于一颗完全二叉树，从根节点到叶子节点，值可以越来
  越小，也可以越来越大。根节点就相当于优先队列的优先值。
  用最小堆实现优先队列，因为只需求第K大的数，每个队列只需存储K个数
  
  31MS	1268K
  */

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define SIZE 1000005

using namespace std;

int heap[SIZE];
int len;

void down(int x) //往下更新
{
    int xx = x * 2;
    int temp = heap[x];
    while(xx <= len) 
    {
        if(xx < len && heap[xx] > heap[xx+1]) //找出左右孩子中最小的那个
            xx ++;
        if(temp <= heap[xx]) //当前值比它的孩子小，堆没有被破坏，无需更新下去
            break;
        else
        {
            heap[x] = heap[xx];
            x = xx;
            xx = x * 2;
        }
    }
    heap[x] = temp;
}

void up(int x) //对新加入的节点要往上更新
{
    int xx = x / 2;
    int temp = heap[x];
    while(xx <= len && temp < heap[xx]) //父亲节点比当前值大，需要更新
    {
        heap[x] = heap[xx];
        x = xx;
        xx = x / 2;
    }
    heap[x] = temp;
}

void insert(int x)
{
    heap[++len] = x;
    up(len);
}

int getMin()
{
    int ret = heap[1];
    return ret;
}

void pop() 
{
    heap[1] = heap[len--];
    down(1);
}

int N,K;
char oper;
int num;

int main()
{
    while(~scanf("%d%d",&N,&K))
    {
        len = 0;
        memset(vis,0,sizeof(vis));
        for(int i=1; i<=N; i++)
        {
            getchar();
            scanf("%c",&oper);
            if(oper == 'I')
            {
                scanf("%d",&num);
                insert(num);
                if(len > K)
                    pop();
            }
            else
            {
                int temp = getMin();
                printf("%d\n",temp);
            }
        }
    }
    return 0;
}