HDU 3282 Running Median 动态中位数，可惜数据范围太小

Running Median

Time Limit: 1 Sec  Memory Limit: 256 MB

题目连接

　　http://acm.hdu.edu.cn/showproblem.php?pid=3282

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

HINT

题意

 

  给你n个数，每次插入一个数，当插入数的数量为奇数的时候，我们就输出中位数

 

题解：

 

这道题要求动态求中位数，但是数据范围太小了，于是我们直接暴力搞就好了

事先排序，然后每次都扫一遍就行了……

 

代码：

 

//qscqesze

#include <cstdio>

#include <cmath>

#include <cstring>

#include <ctime>

#include <iostream>

#include <algorithm>

#include <set>

#include <vector>

#include <sstream>

#include <queue>

#include <typeinfo>

#include <fstream>

#include <map>

typedef long long ll;

using namespace std;

//freopen("D.in","r",stdin);

//freopen("D.out","w",stdout);

#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)

#define maxn 100005

#define mod 10007

#define eps 1e-9

//const int inf=0x7fffffff;   //无限大

const int inf=0x3f3f3f3f;

/*



*/

//**************************************************************************************

inline ll read()

{

    int x=0,f=1;char ch=getchar();

    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}

    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}

    return x*f;

}

struct node

{

    int x,y;

};

node a[maxn];

bool cmp(node b,node c)

{

    return b.x<c.x;

}

int main()

{

    int t=read();

    for(int cas=1;cas<=t;cas++)

    {

        vector<int> ans;

        int n=read(),m=read();

        for(int i=0;i<m;i++)

            a[i].x=read(),a[i].y=i+1;

        sort(a,a+m,cmp);

        for(int i=0;i<m;i+=2)

        {

            int flag=0;

            for(int j=0;j<m;j++)

            {

                if(a[j].y<=i+1)

                    flag++;

                if(flag==(i+2)/2)

                {

                    ans.push_back(a[j].x);

                    break;

                }

            }

        }

        printf("%d %d",cas,ans.size());

        for(int i=0;i<ans.size();i++)

        {

            if((i)%10==0)

                printf("\n");

            if(i%10==0)

                printf("%d",ans[i]);

            else

                printf(" %d",ans[i]);



        }

        printf("\n");



    }

}