【POJ3784】Running Median

Running Median

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 3406		Accepted: 1576

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
解析：
动态维护中位数
方法：
建立两个二叉堆：一个小根堆，一个大根堆。在依次读入这个整数序列的过程中，设当前序列长度为M,我们始终保持：
1、序列中从小到大排名为1~M/2的整数存储在大根堆中：
2、序列中从小到大排名为M/2+1~M的整数存储在小根堆中。
任何时候，如果某一个堆中的元素过多，打破了这个性质，就取出该堆的堆顶插入另一个堆。这样一来，序列的中位数就是小根堆的堆顶。
每次新读入一个数值X后,若X比中位数小，则插入大根堆，否则插入小根堆，在插入之后检查并维护上述性质即可。这就是“对顶堆”算法。
（本题对格式要求严格）

#include<cstring>
#include<cstdio>
#include<algorithm>
#include<vector>
#include<queue>
using namespace std;
int T,n,m,a[50005];

priority_queue<int,vector<int>, greater<int> > q;//从小到大输出：小顶堆 

priority_queue<int> p;//从大到小输出 ：大顶堆 

int main()
{
    scanf("%d",&T);
    while(T--)
    {
        while(!q.empty())q.pop();
        while(!p.empty())p.pop();
        scanf("%d%d",&m,&n);
        printf("%d %d\n",m,(n+1)/2);
        for(int i=1;i<=n;i++) scanf("%d",&a[i]);
        q.push(a[1]);
        printf("%d",a[1]);
        int cnt=1;
        for(int i=2;i<=n;i++)
        {
            if(a[i]>q.top()) q.push(a[i]);
            else p.push(a[i]);
            if(i%2!=0){
                while(p.size()>(i/2))
                {
                    q.push(p.top());
                    p.pop();
                }
                while(q.size()>(i-(i/2)))
                {
                    p.push(q.top());
                    q.pop();
                }
                cnt++;
                if(cnt%10==1) printf("\n%d",q.top());
                else printf(" %d",q.top());
            }    
        }
        puts("");//换行坑人...... 
    }
}

【POJ3784】Running Median

猜你喜欢