POJ3784Running Median
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Running Median
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 3406 | Accepted: 1576 |
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
解析:
动态维护中位数
方法:
建立两个二叉堆:一个小根堆,一个大根堆。在依次读入这个整数序列的过程中,设当前序列长度为M,我们始终保持:
1、序列中从小到大排名为1~M/2的整数存储在大根堆中:
2、序列中从小到大排名为M/2+1~M的整数存储在小根堆中。
任何时候,如果某一个堆中的元素过多,打破了这个性质,就取出该堆的堆顶插入另一个堆。这样一来,序列的中位数就是小根堆的堆顶。
每次新读入一个数值X后,若X比中位数小,则插入大根堆,否则插入小根堆,在插入之后检查并维护上述性质即可。这就是“对顶堆”算法。
(本题对格式要求严格)
#include<cstring> #include<cstdio> #include<algorithm> #include<vector> #include<queue> using namespace std; int T,n,m,a[50005]; priority_queue<int,vector<int>, greater<int> > q;//从小到大输出:小顶堆 priority_queue<int> p;//从大到小输出 :大顶堆 int main() { scanf("%d",&T); while(T--) { while(!q.empty())q.pop(); while(!p.empty())p.pop(); scanf("%d%d",&m,&n); printf("%d %d ",m,(n+1)/2); for(int i=1;i<=n;i++) scanf("%d",&a[i]); q.push(a[1]); printf("%d",a[1]); int cnt=1; for(int i=2;i<=n;i++) { if(a[i]>q.top()) q.push(a[i]); else p.push(a[i]); if(i%2!=0){ while(p.size()>(i/2)) { q.push(p.top()); p.pop(); } while(q.size()>(i-(i/2))) { p.push(q.top()); q.pop(); } cnt++; if(cnt%10==1) printf(" %d",q.top()); else printf(" %d",q.top()); } } puts("");//换行坑人...... } }
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