机试练习08：poj3784——动态堆求解中位数

Posted 2020-10-30 Alyssa_young

一、题解方法

建立一个最小堆和一个最大堆，不断更新当前中位数，建立一个数组存入读入奇数个数时的中位数。

用stl中的优先队列存放最大堆、最小堆。

二、题解代码

 1 #include "stdio.h"
 2 #include "stdlib.h"
 3 #include <iostream>
 4 #include <queue>
 5 #include <algorithm>
 6 
 7 using namespace std;
 8 
 9 const int MAX = 10000;
10 const int ACCOUNT = 10;
11 priority_queue<int, vector<int>, greater<int> > l;
12 priority_queue<int, vector<int>, less<int> > g;
13 
14 int ans[MAX];
15 int p;
16 
17 int main()
18 {
19     int case_num;
20     scanf("%d", &case_num);
21     int i;
22     while(case_num--)
23     {
24         int case_order, num, mid;
25         scanf("%d%d%d", &case_order, &num, &mid);
26         int middle_num = (num+1) / 2;
27         printf("%d %d\n", case_order, middle_num);
28         p = 0;
29         ans[p++] = mid;
30         while ( !l.empty() ) l.pop();
31         while ( !g.empty() ) g.pop();
32         
33         for(i = 2; i <= num; i++)
34         {
35             int temp;
36             scanf("%d", &temp);
37             if ( temp < mid )
38             {
39                 g.push(temp);
40                 if ( g.size() - l.size() == 2 )
41                 {
42                     l.push(mid);
43                     mid = g.top();
44                     g.pop();
45                 }
46             }
47             else
48             {
49                 l.push(temp);
50                 if(l.size() - g.size() == 2)
51                 {
52                     g.push(mid);
53                     mid = l.top();
54                     l.pop();
55                 }
56             }
57             if ( i & 1 )
58             {
59                 ans[p++] = mid;
60             }
61         }
62         for(i = 0; i < p; i++)
63         {
64             printf("%d", ans[i]);
65             if(i % ACCOUNT == 9 || i == p-1)
66                 printf("\n");
67             else
68                 printf(" ");
69         }
70         
71         
72     }
73     return 0;
74 }