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poj 3784（对顶堆）
Running Median
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 1824 | Accepted: 889 |
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 51 2 3 4 52 59 8 7 6 53 1223 23 22 22 13 3 5 5 3 -3 -7 -3
题意
给n个数，第奇数次输入的时候，输出当前数中的中位数。
思路
对顶堆。建立一个大根堆和小根堆，如果当前元素大于小根堆堆顶元素则放入小根堆，否则放入大根堆。
这样保证小根堆的任意元素>大根堆的任意元素，并且mnq.size==mxq.size+1，即小根堆比大根堆多1个元素（小根堆堆顶元素:中位数）
如果不满足上面的等式，则对两个堆进行调整。
1 #include <iostream> 2 #include <cstdio> 3 #include <queue> 4 #include <functional> 5 #include <vector> 6 7 using namespace std; 8 9 priority_queue<int> mxq;10 priority_queue<int,vector<int>,greater<int> > mnq;11 12 vector<int> res;13 14 void add(int x){15 if(mnq.empty()){16 mnq.push(x);17 return;18 }19 if(x>mnq.top()) mnq.push(x);20 else mxq.push(x);21 while(mnq.size()<mxq.size()){22 mnq.push(mxq.top());23 mxq.pop();24 }25 while(mnq.size()>mxq.size()+1){26 mxq.push(mnq.top());27 mnq.pop();28 }29 }30 31 int main(){32 int T;33 scanf("%d",&T);34 while(T--){35 while(!mnq.empty()) mnq.pop();36 while(!mxq.empty()) mxq.pop();37 res.clear();38 int n,m;39 scanf("%d %d",&n,&m);40 for(int i=1;i<=m;i++){41 int x;scanf("%d",&x);42 add(x);43 if(i%2) res.push_back(mnq.top());44 }45 printf("%d %d\n",n,res.size());46 for(int i=0;i<res.size();i++){47 if(i%10==0&&i) putchar(‘\n‘);48 if(i%10>=1) putchar(‘ ‘);49 printf("%d",res[i]);50 }51 printf("\n");52 }53 return 0;54 }
poj 3784（对顶堆）
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