POJ3784 Running Median 题解

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题目描述
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.
输入
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.
输出
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.
样例输入

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

样例输出

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

来源
Greater New York Regional 2009

题目请自己前往谷歌生草机翻译

题目解析：一道动态中位数，就是输出比较玄学。

方法一：平衡树
这里不做介绍，因为就是板子。

方法二：堆
建立两个堆，对这两个堆进行操作。一个堆负责大于等于中位数的数的存贮，使用小根堆；另一个堆负责小于等于中位数的数的存贮，使用大根堆。
无疑，每次插入的时候，都要和中位数比较，比中位数大的进入小根堆，比中位数小的进入大根堆。
如果大根堆的数字比小根堆的数字多 \(2\) 那么将大根堆的一个数字进入小根堆，反之亦然。
查询的时候只要比较哪个堆的数字多，中位数就是数字多的堆的堆顶。
如果数字为偶数，我们就默认中位数为两个堆的中位数的平均值。

代码

#include<cstdio>
#include<queue>
#include<vector>
using namespace std;
inline int read(){
	char c=getchar();
	int sum=0,flag=0;
	while((c<‘0‘||c>‘9‘)&&c!=‘-‘) c=getchar();
	if(c==‘-‘)
		c=getchar(),flag=1;
	while(‘0‘<=c&&c<=‘9‘){
		sum=(sum<<1)+(sum<<3)+(c^48);
		c=getchar();
	}
	if(flag) return -sum;
	return sum;
}
priority_queue<int> big;
priority_queue< int , vector<int> , greater<int> > small;
int n,x,T,mid,k,a;
int main(){
	//freopen("1.in","r",stdin);
    T=read();
    while(T--){
    	while(!big.empty()) big.pop();
    	while(!small.empty()) small.pop();
    	k=read();
		n=read();
    	x=read();
    	mid=x;
    	printf("%d %d\n",k,n+1>>1);
    	big.push(x);
		printf("%d ",x);
		a=1;
		for(int i=2;i<=n;i++){
    		x=read();
    		if(x>mid) small.push(x);
    		else big.push(x);
    		if(big.size()>small.size()+1){
    			small.push(big.top());
    			big.pop();
			}
			else if(big.size()+1<small.size()){
				big.push(small.top());
				small.pop();
			}
			if(i&1){
				if(big.size()>small.size()){
					mid=big.top();
					a++;
					if(a%10==0) printf("%d\n",mid);
					else printf("%d ",mid);
				}
				else{
					a++;
					mid=small.top();
					if(a%10==0) printf("%d\n",mid);
					else printf("%d ",mid);
				}
			}
			else{
				mid=big.top()+small.top()>>1;
			}
		}
		if(T) printf("\n");
	}
	return 0;
}

POJ3784 Running Median 题解

标签：std   etc   ast   ==   top   ble   sum   HERE   turn   

原文地址：https://www.cnblogs.com/jiangtaizhe001/p/14405556.html