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

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

维护两个堆;

一个大根堆,一个小根堆。

 1 /*
 2     problem:poj 3784
 3     by:S.B.S.
 4 */
 5 #include<iostream>
 6 #include<cstdio>
 7 #include<cstring>
 8 #include<cmath>
 9 #include<algorithm>
10 #include<queue>
11 #include<cstdlib>
12 #include<iomanip>
13 #include<cassert>
14 #include<climits>
15 #define maxn 10001
16 #define F(i,j,k) for(int i=j;i<=k;i++)
17 #define M(a,b) memset(a,b,sizeof(a))
18 #define FF(i,j,k) for(int i=j;i>=k;i--)
19 #define inf 0x7fffffff
20 #define p 23333333333333333
21 using namespace std;
22 int read(){
23     int x=0,f=1;char ch=getchar();
24     while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
25     while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
26     return x*f;
27 }
28 priority_queue<int,vector<int>,greater<int> > q1;
29 priority_queue<int,vector<int>,less<int> > q2;
30 vector<int> g;
31 void add(int x)
32 {
33     if(q1.empty()){
34         q1.push(x);
35         return;
36     }
37     if(x>q1.top()) q1.push(x);
38     else q2.push(x);
39     while(q1.size()<q2.size()){
40         q1.push(q2.top());
41         q2.pop();
42     }
43     while(q1.size()>q2.size()+1) {
44         q2.push(q1.top());
45         q1.pop();
46     }
47 }
48 int main()
49 {
50     std::ios::sync_with_stdio(false);//cout<<setiosflags(ios::fixed)<<setprecision(1)<<y;
51 //  freopen("data.in","r",stdin);
52 //  freopen("data.out","w",stdout);
53     int t,c,n,x;
54     cin>>t;
55     while(t--){
56         while(!q1.empty()) q1.pop();
57         while(!q2.empty()) q2.pop();
58         g.clear();
59         cin>>c>>n;
60         F(i,0,n-1){
61             cin>>x;
62             add(x);
63             if(i%2==0) g.push_back(q1.top());
64         }
65         cout<<c<<" "<< (n+1)/2 <<endl;
66         F(i,0,g.size()-1){
67             if(i>0&& i%10==0) cout<<endl;
68             if(i%10) cout<<" ";
69             cout<<g[i];
70         }
71         cout<<endl;
72     }
73     return 0;
74 }
poj 3784

 

posted @ 2016-07-06 22:50  SBSOI  阅读(252)  评论(0编辑  收藏  举报