poj 3784（对顶堆）

Running Median

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 1824		Accepted: 889

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


题意
给n个数，第奇数次输入的时候，输出当前数中的中位数。


思路
对顶堆。建立一个大根堆和小根堆，如果当前元素大于小根堆堆顶元素则放入小根堆，否则放入大根堆。
这样保证小根堆的任意元素>大根堆的任意元素，并且mnq.size==mxq.size+1，即小根堆比大根堆多1个元素（小根堆堆顶元素:中位数）
如果不满足上面的等式，则对两个堆进行调整。

#include <iostream>
#include <cstdio>
#include <queue>
#include <functional>
#include <vector>

using namespace std;

priority_queue<int> mxq;
priority_queue<int,vector<int>,greater<int> > mnq;

vector<int> res;

void add(int x){
    if(mnq.empty()){
        mnq.push(x);
        return;
    }
    if(x>mnq.top()) mnq.push(x);
    else mxq.push(x);
    while(mnq.size()<mxq.size()){
        mnq.push(mxq.top());
        mxq.pop();
    }
    while(mnq.size()>mxq.size()+1){
        mxq.push(mnq.top());
        mnq.pop();
    }
}

int main(){
    int T;
    scanf("%d",&T);
    while(T--){
        while(!mnq.empty()) mnq.pop();
        while(!mxq.empty()) mxq.pop();
        res.clear();
        int n,m;
        scanf("%d %d",&n,&m);
        for(int i=1;i<=m;i++){
            int x;scanf("%d",&x);
            add(x);
            if(i%2) res.push_back(mnq.top());
        }
        printf("%d %d
",n,res.size());
        for(int i=0;i<res.size();i++){
            if(i%10==0&&i) putchar('
');
            if(i%10>=1) putchar(' ');
            printf("%d",res[i]);
        }
        printf("
");
    }
    return 0;
}