实验6选第k小元素：特定分治策略
作者：互联网
问题：
给出一个数组$a[n]$，求第$k$小元素是什么。
解析：
分治思想，将数组五个一组划分，并计算出每组数的中位数。然后把各组中位数的中位数找出。统计数组中小于中位数的个数$num$，有三种情况。
① $num = k$，则中位数就是要查询的数。
② $num > k$，则在小于中位数的集合中查询第$k$小。
③ $num < k$，则在大于中位数的集合中查询第$k – num$小。
设计（核心代码）：
1 void insertsort(int a[], int l, int r)//从小到大排序
2 {
3 int i, j, key;
4 for (i = l + 1; i <= r; ++i)
5 {
6 key = a[i];
7 for (j = i - 1; j >= l && key < a[j]; --j)
8 {
9 a[j + 1] = a[j];
10 }
11 a[j + 1] = key;
12 }
13 }
14
15 int partition(int a[], int l, int r, int pivot)
16 {
17 int x, i = l - 1, j;
18 for (j = l; j < r; ++j)
19 {
20 if (a[j] == pivot) swap(a[j], a[r]);
21 }
22 x = a[r];
23 for (j = l; j < r; ++j)
24 {
25 if (a[j] <= x)
26 {
27 ++i;
28 swap(a[i], a[j]);
29 }
30 }
31 swap(a[r], a[i + 1]);
32 return i + 1;
33 }
34
35 int select(int a[], int l, int r, int k)
36 {
37 int group, i, left, right, mid;
38 int pivot, p, lnum;
39 if (r - l + 1 <= 5)
40 {
41 insertsort(a, l, r);
42 return a[l + k - 1];
43 }
44 group = (r - l + 1 + 5) / 5;
45 for (i = 0; i < group; ++i)
46 {
47 left = l + 5 * i;
48 right = (l + 5 * i + 4) > r ? r : l + 5 * i + 4;
49 mid = (left + right) / 2;
50 insertsort(a, left, right);
51 swap(a[l + i], a[mid]);
52 }
53 pivot = select(a, l, l + group - 1, (group + 1) / 2);
54 p = partition(a, l, r, pivot);
55 lnum = p - l;
56 if (k == lnum + 1)
57 return a[p];
58 else if (k <= lnum)
59 return select(a, l, p - 1, k);
60 else
61 return select(a, p + 1, r, k - lnum - 1);
62 }
分析：
复杂度：$O(n)$。
源码：
https://github.com/Big-Kelly/Algorithm
1 #include<bits/stdc++.h>
2 #include <set>
3 #include <map>
4 #include <stack>
5 #include <cmath>
6 #include <queue>
7 #include <cstdio>
8 #include <string>
9 #include <vector>
10 #include <cstring>
11 #include <iostream>
12 #include <algorithm>
13
14 #define ll long long
15 #define pll pair<ll,ll>
16 #define pii pair<int,int>
17 #define bug printf("*********\n")
18 #define FIN freopen("input.txt","r",stdin);
19 #define FON freopen("output.txt","w+",stdout);
20 #define IO ios::sync_with_stdio(false),cin.tie(0)
21 #define ls root<<1
22 #define rs root<<1|1
23 #define Q(a) cout<<a<<endl
24
25 using namespace std;
26 const int inf = 2e9 + 7;
27 const ll Inf = 1e18 + 7;
28 const int maxn = 1e6 + 5;
29 const int mod = 1e9 + 7;
30
31 ll gcd(ll a, ll b)
32 {
33 return b ? gcd(b, a % b) : a;
34 }
35
36 ll lcm(ll a, ll b)
37 {
38 return a / gcd(a, b) * b;
39 }
40
41 ll read()
42 {
43 ll p = 0, sum = 0;
44 char ch;
45 ch = getchar();
46 while (1)
47 {
48 if (ch == '-' || (ch >= '0' && ch <= '9'))
49 break;
50 ch = getchar();
51 }
52
53 if (ch == '-')
54 {
55 p = 1;
56 ch = getchar();
57 }
58 while (ch >= '0' && ch <= '9')
59 {
60 sum = sum * 10 + ch - '0';
61 ch = getchar();
62 }
63 return p ? -sum : sum;
64 }
65
66 void insertsort(int a[], int l, int r)//从小到大排序
67 {
68 int i, j, key;
69 for (i = l + 1; i <= r; ++i)
70 {
71 key = a[i];
72 for (j = i - 1; j >= l && key < a[j]; --j)
73 {
74 a[j + 1] = a[j];
75 }
76 a[j + 1] = key;
77 }
78 }
79
80 int partition(int a[], int l, int r, int pivot)
81 {
82 int x, i = l - 1, j;
83 for (j = l; j < r; ++j)
84 {
85 if (a[j] == pivot) swap(a[j], a[r]);
86 }
87 x = a[r];
88 for (j = l; j < r; ++j)
89 {
90 if (a[j] <= x)
91 {
92 ++i;
93 swap(a[i], a[j]);
94 }
95 }
96 swap(a[r], a[i + 1]);
97 return i + 1;
98 }
99
100 int select(int a[], int l, int r, int k)
101 {
102 int group, i, left, right, mid;
103 int pivot, p, lnum;
104 if (r - l + 1 <= 5)
105 {
106 insertsort(a, l, r);
107 return a[l + k - 1];
108 }
109 group = (r - l + 1 + 5) / 5;
110 for (i = 0; i < group; ++i)
111 {
112 left = l + 5 * i;
113 right = (l + 5 * i + 4) > r ? r : l + 5 * i + 4;
114 mid = (left + right) / 2;
115 insertsort(a, left, right);
116 swap(a[l + i], a[mid]);
117 }
118 pivot = select(a, l, l + group - 1, (group + 1) / 2);
119 p = partition(a, l, r, pivot);
120 lnum = p - l;
121 if (k == lnum + 1)
122 return a[p];
123 else if (k <= lnum)
124 return select(a, l, p - 1, k);
125 else
126 return select(a, p + 1, r, k - lnum - 1);
127 }
128
129 int a[maxn];
130 int n, k;
131
132 int main()
133 {
134 scanf("%d %d", &n, &k);
135 for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
136 printf("%d\n", select(a, 1, n, k));
137 }
View Code
标签：分治,int,元素,中位数,num,选第,pivot,include,define 来源： https://www.cnblogs.com/zhang-Kelly/p/12656365.html