POJ 3784 Running Median

最新推荐文章于 2021-11-21 15:10:34 发布

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最新推荐文章于 2021-11-21 15:10:34 发布

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本文介绍了一种通过使用两个堆（大根堆和小根堆）来高效维护一系列整数中位数的方法，并提供了一个完整的C++实现示例。该算法能够确保在O(nlogn)的时间复杂度下，每读取一个奇数位置的整数时输出当前序列的中位数。

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Description

For this problem, you will write a program that reads in a sequence of 32-bit signed integers. After each odd-indexed value is read, output the median (middle value) of the elements received so far.

Input

The first line of input contains a single integer P, (1 ≤ P ≤ 1000), which is the number of data sets that follow. The first line of each data set contains the data set number, followed by a space, followed by an odd decimal integer M, (1 ≤ M ≤ 9999), giving the total number of signed integers to be processed. The remaining line(s) in the dataset consists of the values, 10 per line, separated by a single space. The last line in the dataset may contain less than 10 values.

Output

For each data set the first line of output contains the data set number, a single space and the number of medians output (which should be one-half the number of input values plus one). The output medians will be on the following lines, 10 per line separated by a single space. The last line may have less than 10 elements, but at least 1 element. There should be no blank lines in the output.

Sample Input
3
1 9
1 2 3 4 5 6 7 8 9
2 9
9 8 7 6 5 4 3 2 1
3 23
23 41 13 22 -3 24 -31 -11 -8 -7
3 5 103 211 -311 -45 -67 -73 -81 -99
-33 24 56

Sample Output
1 5
1 2 3 4 5
2 5
9 8 7 6 5
3 12
23 23 22 22 13 3 5 5 3 -3
-7 -3

题目大意：依次添加 n 个数，要求第奇数次添加时，输出当前数列中的中位数。
然而，POJ的输入输出总是十分奇葩。。。
维护两个堆，一个大根堆，一个小根堆显然应该让大根堆存较小的数，小根堆存较大的数如果维护好了这两个堆，令当前的中位数就是小根堆的堆顶每次新加入一个数，如果这个数比当前的中位数大，就存入小根堆，否则存入大根堆限制小根堆中元素的个数比大根堆多 1 或相等如果个数相差太多，则将元素多的堆删掉堆顶插入到元素少的堆中此时就可以保证当前的中位数就是小根堆的堆顶时间复杂度 O(nlogn)

#include<iostream>
#include<cstdio>
#include<queue>
using namespace std;
priority_queue<int>bg;
priority_queue< int,vector<int>,greater<int> >sm;
int main()
{
    int q,x;
    scanf("%d",&q);
    while(q--)
    {
        int d,n,t=0;
        scanf("%d%d",&d,&n);
        printf("%d %d\n",d,(n+1)/2);
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&x);
            if(sm.empty()||x>sm.top())
                sm.push(x);
            else
                bg.push(x);
            if(i%2==1)
            {
                while(sm.size()>bg.size()+1)
                {
                    x=sm.top();
                    sm.pop();
                    bg.push(x);
                }
                while(sm.size()<bg.size()+1)
                {
                    x=bg.top();
                    bg.pop();
                    sm.push(x);
                }
                t++;
                if(t%10!=0)
                    printf("%d ",sm.top());
                else
                    printf("%d\n",sm.top());
            }
        }
        if(((n+1)/2)%10!=0)
            printf("\n");
        while(!sm.empty())
            sm.pop();
        while(!bg.empty())
            bg.pop();
    }
    return 0;
}