用二分法求众数

一、题目引发：

Boxing Packing

Description:

Mishka has got n empty boxes. For every i (1 ≤ i ≤ n), i-th box is a cube with side length a_i.

Mishka can put a box i into another box j if the following conditions are met:

• i-th box is not put into another box;
• j-th box doesn't contain any other boxes;
• box i is smaller than box j (a_i < a_j).

Mishka can put boxes into each other an arbitrary number of times. He wants to minimize the number of visible boxes. A box is called visible iff it is not put into some another box.

Help Mishka to determine the minimum possible number of visible boxes!

理解一下就是：给定一串序列，至少要拆分成多少组互异的子序列。

进一步理解：等价于求众数的重数。

二、分治法求众数及其重数

算法描述：首先将序列排好序，取中间值，分别左右遍历得到最长连续子序列，设此长度为L。若左子区间长度大于L,则向左递归；若右子区间长度大于L,则向右递归；

算法实现：

代码：

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<stdlib.h>
const int maxn = 5000 + 10;
int len[maxn];
int targrt,cnt = -1;

int compare(const void * a, const void * b)
{
    return (*(int*)a - *(int*)b);
}
void split(int * len, int l, int r)
{
    int mid = (l + r) / 2;
    int i, j;
    for (i = l; i <= mid; i++)
    {
        if (len[i] == len[mid])
            break;
    }
    for (j = mid + 1; j <= r; j++)
    {
        if (len[j] != len[mid])
            break;
    }
    if ((j - i) > cnt)
    {
        cnt = j - i;
        targrt = len[mid];
    }
    if ((i - l) > cnt)
        split(len, l, i - 1);
    if ((r - j + 1) > cnt)
        split(len, j, r);
    return;
}
int main()
{
    int n;
    scanf("%d
", &n);
    for (int i = 0; i < n; i++)
        scanf("%d", &len[i]);
    qsort(len, n, sizeof(int),compare);
    split(len, 0, n-1);
    printf("%d",cnt);
    return 0;
}

三、其他方法

还可以参照“桶”的思想，每个数字就是桶的编号，桶的内容就是该数的重数。但是这个数字可能很大，就要采用哈希的办法，这个方法以后再补充吧！

而且这题不用二分法也能过，采用依次遍历，时间复杂度为O(N)（二分法的时间复杂度为O(log(N))）,代码实现起来非常的简单

代码：

#include<stdio.h>
#include<iostream>
#include<algorithm>
using namespace std;
const int maxn = 5000 + 10;
int a[maxn];

int main()
{
    int n;
    scanf("%d", &n);
    for (int i = 0; i < n; i++)
        scanf("%d", &a[i]);
    sort(a , a + n);
    int ans = 1, cnt = 1;
    for (int i = 1; i < n; i++)
    {
        if (a[i] == a[i - 1])
        {
            cnt++;
            if (cnt > ans)
                ans = cnt;
        }
        else
            cnt = 1;
    }
    printf("%d
", ans);
}