POJ 3784 Running Median

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

【分析】
二叉堆维护
分别维护一个大根堆和一个小根堆。
操作：如果比当前中位数k大的话那么进小根堆，反之进大根堆
维护：小根堆元素比大根堆元素多一个
具体做法：两个两个的加元素
用到了C++的STL，很省代码长度

【代码】

//poj 3784 Running Median#include<iostream>#include<cstdio>#include<queue>#include<algorithm>#define fo(i,j,k) for(i=j;i<=k;i++)using namespace std;priority_queue <int> da;priority_queue < int,vector<int>,greater<int> > xiao;int Q,n,x,y,t;int a[5001];bool b;int main(){    int i,j,k,p,f,d;   //k为中位数     scanf("%d",&Q);    while(Q--)    {        t=0;b=0;        scanf("%d%d%d",&d,&n,&x);        if(n%2==0) b=1;        k=x;        a[++t]=k;        xiao.push(k);        fo(i,1,(n-1)/2)        {            scanf("%d%d",&x,&y);            if(x>=k) xiao.push(x);            else da.push(x);            if(y>=k) xiao.push(y);            else da.push(y);            while(xiao.size()!=da.size()+1 && xiao.size()>da.size())             //如果小根堆大             {                p=xiao.top();                xiao.pop();                da.push(p);             }            while(xiao.size()!=da.size()+1 && da.size()>xiao.size())            //如果大根堆大             {                p=da.top();                da.pop();                xiao.push(p);             }            k=xiao.top();            a[++t]=k;        }        if(b) scanf("%d",&x);        printf("%d %d\n",d,(n+1)>>1);        fo(i,1,t)        {            printf("%d ",a[i]);            if(i%10==0) printf("\n");        }        printf("\n");        while(!xiao.empty()) xiao.pop();        while(!da.empty()) da.pop();    }    return 0;}