POJ 3784 Running Median

Running Median

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 2024		Accepted: 972

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

思路：对于输入的前2k个数，用一个大根堆存较小的k个数，一个小根堆存较大的k个数，每次插入元素时，与堆顶比较就行了。这个东西叫对顶堆，又可称为中根堆。复杂度nlogn。

#include <iostream>
#include <fstream>
#include <sstream>
#include <cstdlib>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <queue>
#include <stack>
#include <vector>
#include <set>
#include <map>
#include <list>
#include <iomanip>
#include <cctype>
#include <cassert>
#include <bitset>
#include <ctime>

using namespace std;

#define pau system("pause")
#define ll long long
#define pii pair<int, int>
#define pb push_back
#define mp make_pair
#define clr(a, x) memset(a, x, sizeof(a))

const double pi = acos(-1.0);
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const double EPS = 1e-9;

int T, n, x, ca;
vector<int> ans;
priority_queue<int, vector<int>, less<int> > que1;
priority_queue<int, vector<int>, greater<int> > que2;
int main() {
    scanf("%d", &T);
    while (T--) {
        ans.clear();
        while (que1.size()) que1.pop();
        while (que2.size()) que2.pop();
        scanf("%d%d", &ca, &n);
        for (int i = 1; i <= n; ++i) {
            scanf("%d", &x);
            if (que1.size() < que2.size()) {
                if (que2.size() && x > que2.top()) {
                    que2.push(x);
                    x = que2.top();
                    que2.pop();
                }
                que1.push(x);
            } else if (que1.size() > que2.size()) {
                if (que1.size() && x < que1.top()) {
                    que1.push(x);
                    x = que1.top();
                    que1.pop();
                }
                que2.push(x);
            } else {
                if (que1.empty() || x < que1.top()) {
                    que1.push(x);
                    ans.pb(que1.top());
                } else {
                    que2.push(x);
                    ans.pb(que2.top());
                }
            }
        }
        printf("%d %d
", ca, ans.size());
        for (int i = 0; i < ans.size(); ++i) {
            printf("%d", ans[i]);
            putchar(9 == i % 10 ? '
' : ' ');
        }
        if (ans.size() % 10) puts("");
    }
    return 0;
}