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
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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 2 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; }