实验6选第k小元素：特定分治策略

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）：

void insertsort(int a[], int l, int r)//从小到大排序
{
    int i, j, key;
    for (i = l + 1; i <= r; ++i)
    {
        key = a[i];
        for (j = i - 1; j >= l && key < a[j]; --j)
        {
            a[j + 1] = a[j];
        }
        a[j + 1] = key;
    }
}

int partition(int a[], int l, int r, int pivot)
{
    int x, i = l - 1, j;
    for (j = l; j < r; ++j)
    {
        if (a[j] == pivot)    swap(a[j], a[r]);
    }
    x = a[r];
    for (j = l; j < r; ++j)
    {
        if (a[j] <= x)
        {
            ++i;
            swap(a[i], a[j]);
        }
    }
    swap(a[r], a[i + 1]);
    return i + 1;
}

int select(int a[], int l, int r, int k)
{
    int group, i, left, right, mid;
    int pivot, p, lnum;
    if (r - l + 1 <= 5)
    {
        insertsort(a, l, r);
        return a[l + k - 1];
    }
    group = (r - l + 1 + 5) / 5;
    for (i = 0; i < group; ++i)
    {
        left = l + 5 * i;
        right = (l + 5 * i + 4) > r ? r : l + 5 * i + 4;
        mid = (left + right) / 2;
        insertsort(a, left, right);
        swap(a[l + i], a[mid]);
    }
    pivot = select(a, l, l + group - 1, (group + 1) / 2);
    p = partition(a, l, r, pivot);
    lnum = p - l;
    if (k == lnum + 1)
        return a[p];
    else if (k <= lnum)
        return select(a, l, p - 1, k);
    else
        return select(a, p + 1, r, k - lnum - 1);
}

分析：

复杂度：$O(n)$。

源码：

https://github.com/Big-Kelly/Algorithm

#include<bits/stdc++.h>
#include <set>
#include <map>
#include <stack>
#include <cmath>
#include <queue>
#include <cstdio>
#include <string>
#include <vector>
#include <cstring>
#include <iostream>
#include <algorithm>

#define ll long long
#define pll pair<ll,ll>
#define pii pair<int,int>
#define bug printf("*********
")
#define FIN freopen("input.txt","r",stdin);
#define FON freopen("output.txt","w+",stdout);
#define IO ios::sync_with_stdio(false),cin.tie(0)
#define ls root<<1
#define rs root<<1|1
#define Q(a) cout<<a<<endl

using namespace std;
const int inf = 2e9 + 7;
const ll Inf = 1e18 + 7;
const int maxn = 1e6 + 5;
const int mod = 1e9 + 7;

ll gcd(ll a, ll b)
{
    return b ? gcd(b, a % b) : a;
}

ll lcm(ll a, ll b)
{
    return a / gcd(a, b) * b;
}

ll read()
{
    ll p = 0, sum = 0;
    char ch;
    ch = getchar();
    while (1)
    {
        if (ch == '-' || (ch >= '0' && ch <= '9'))
            break;
        ch = getchar();
    }

    if (ch == '-')
    {
        p = 1;
        ch = getchar();
    }
    while (ch >= '0' && ch <= '9')
    {
        sum = sum * 10 + ch - '0';
        ch = getchar();
    }
    return p ? -sum : sum;
}

void insertsort(int a[], int l, int r)//从小到大排序
{
    int i, j, key;
    for (i = l + 1; i <= r; ++i)
    {
        key = a[i];
        for (j = i - 1; j >= l && key < a[j]; --j)
        {
            a[j + 1] = a[j];
        }
        a[j + 1] = key;
    }
}

int partition(int a[], int l, int r, int pivot)
{
    int x, i = l - 1, j;
    for (j = l; j < r; ++j)
    {
        if (a[j] == pivot)    swap(a[j], a[r]);
    }
    x = a[r];
    for (j = l; j < r; ++j)
    {
        if (a[j] <= x)
        {
            ++i;
            swap(a[i], a[j]);
        }
    }
    swap(a[r], a[i + 1]);
    return i + 1;
}

int select(int a[], int l, int r, int k)
{
    int group, i, left, right, mid;
    int pivot, p, lnum;
    if (r - l + 1 <= 5)
    {
        insertsort(a, l, r);
        return a[l + k - 1];
    }
    group = (r - l + 1 + 5) / 5;
    for (i = 0; i < group; ++i)
    {
        left = l + 5 * i;
        right = (l + 5 * i + 4) > r ? r : l + 5 * i + 4;
        mid = (left + right) / 2;
        insertsort(a, left, right);
        swap(a[l + i], a[mid]);
    }
    pivot = select(a, l, l + group - 1, (group + 1) / 2);
    p = partition(a, l, r, pivot);
    lnum = p - l;
    if (k == lnum + 1)
        return a[p];
    else if (k <= lnum)
        return select(a, l, p - 1, k);
    else
        return select(a, p + 1, r, k - lnum - 1);
}

int a[maxn];
int n, k;

int main()
{
    scanf("%d %d", &n, &k);
    for (int i = 1; i <= n; ++i)    scanf("%d", &a[i]);
    printf("%d
", select(a, 1, n, k));
}