poj 3784 Running Median

Running Median

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 1652		Accepted: 818

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

Source

Greater New York Regional 2009

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————————————————————我是分割线——————————————————————————

维护两个堆；

一个大根堆，一个小根堆。

/*
    problem:poj 3784
    by:S.B.S.
*/
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<queue>
#include<cstdlib>
#include<iomanip>
#include<cassert>
#include<climits>
#define maxn 10001
#define F(i,j,k) for(int i=j;i<=k;i++)
#define M(a,b) memset(a,b,sizeof(a))
#define FF(i,j,k) for(int i=j;i>=k;i--)
#define inf 0x7fffffff
#define p 23333333333333333
using namespace std;
int 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;
}
priority_queue<int,vector<int>,greater<int> > q1;
priority_queue<int,vector<int>,less<int> > q2;
vector<int> g;
void add(int x)
{
    if(q1.empty()){
        q1.push(x);
        return;
    }
    if(x>q1.top()) q1.push(x);
    else q2.push(x);
    while(q1.size()<q2.size()){
        q1.push(q2.top());
        q2.pop();
    }
    while(q1.size()>q2.size()+1) {
        q2.push(q1.top());
        q1.pop();
    }
}
int main()
{
    std::ios::sync_with_stdio(false);//cout<<setiosflags(ios::fixed)<<setprecision(1)<<y;
//  freopen("data.in","r",stdin);
//  freopen("data.out","w",stdout);
    int t,c,n,x;
    cin>>t;
    while(t--){
        while(!q1.empty()) q1.pop();
        while(!q2.empty()) q2.pop();
        g.clear();
        cin>>c>>n;
        F(i,0,n-1){
            cin>>x;
            add(x);
            if(i%2==0) g.push_back(q1.top());
        }
        cout<<c<<" "<< (n+1)/2 <<endl;
        F(i,0,g.size()-1){
            if(i>0&& i%10==0) cout<<endl;
            if(i%10) cout<<" ";
            cout<<g[i];
        }
        cout<<endl;
    }
    return 0;
}