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
题目大意：
输⼊⼀个序列。
请分别求出前1、3、5、7、9……个数字的中位数
解题思路：
维护一个大根堆和一个小根堆，让他们的⼤⼩保持⾄多差1。
如果⼤⼩差距⼤于1，取出较⼤的堆的⼀个堆顶放到另⼀堆⾥即可。
询问中位数时，较⼤的堆的堆顶就是答案。
感谢PoPoQQQ简明易懂的配图
AC代码：
1 #include<iostream> 2 #include<cstring> 3 #include<queue> 4 using namespace std; 5 int p,m,k,pp,mid,ans[100000]; 6 priority_queue<int> a;//大根堆 7 priority_queue<int ,vector<int>, greater<int> > b;//小根堆 8 int main() { 9 cin >> p; 10 for(int i = 1; i <= p; i++) { 11 cin >> pp >> m; 12 int xb = 0; 13 memset(ans, 0,sizeof(ans)); 14 while(!a.empty()) a.pop();//初始化 15 while(!b.empty()) b.pop();//初始化 16 for(int x = 1;x <= m; ++x) { 17 cin >> k; 18 if(x==1) b.push(k); 19 else{ 20 if(k > b.top()) b.push(k);//如果比中位数大，就扔到小根堆里 21 else if(k < b.top()) a.push(k);//如果比中位数小 ，就扔到大根堆里 22 if(k == b.top()) {//如果等于中位数，就放到元素较少的堆里 23 if(a.size() > b.size()) b.push(k); 24 else a.push(k); 25 } 26 } 27 28 if(a.size() >= b.size() + 2) {//保持两个堆元素数量相差小于二 29 b.push(a.top()); 30 a.pop(); 31 mid = a.top(); 32 } 33 else if(a.size() + 2 <= b.size()) {//保持两个堆元素数量相差小于二 34 a.push(b.top()); 35 b.pop(); 36 mid = b.top(); 37 } 38 if(a.size() > b.size()) mid = a.top(); 39 if(b.size() > a.size()) mid = b.top();//中位数为元素较多的堆的堆顶 40 if(x % 2 == 1) { 41 xb++; 42 ans[xb] = mid; 43 } 44 45 } 46 cout << pp << ' ' << xb <<endl; 47 for(int x = 1;x <= xb;++x){ 48 if(x > 10 && x % 10 == 1) cout << endl; 49 cout << ans[x] << " "; 50 } 51 cout << endl; 52 } 53 54 55 return 0; 56 }