Not so Diverse（AtCoder-3719）

最新推荐文章于 2022-06-28 21:27:42 发布

Alex_McAvoy

最新推荐文章于 2022-06-28 21:27:42 发布

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分类专栏： # AtCoder # 常用技巧——桶排

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常用技巧——桶排

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本文探讨了在有限重写次数下，使球上序号种类不超过特定数量的算法问题。通过桶排序统计不同序号出现频率，实现最小化重写次数的目标。

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Problem Description

Takahashi has N balls. Initially, an integer Ai is written on the i-th ball.

He would like to rewrite the integer on some balls so that there are at most K different integers written on the N balls.

Find the minimum number of balls that Takahashi needs to rewrite the integers on them.

Constraints

1≤K≤N≤200000
1≤Ai≤N
All input values are integers.
Input

Input is given from Standard Input in the following format:

N K
A1 A2 ... AN

Output

Print the minimum number of balls that Takahashi needs to rewrite the integers on them.

Example

Sample Input 1

5 2
1 1 2 2 5

Sample Output 1

1
For example, if we rewrite the integer on the fifth ball to 2, there are two different integers written on the balls: 1 and 2. On the other hand, it is not possible to rewrite the integers on zero balls so that there are at most two different integers written on the balls, so we should print 1.

Sample Input 2

4 4
1 1 2 2

Sample Output 2

0
Already in the beginning, there are two different integers written on the balls, so we do not need to rewrite anything.

Sample Input 3

10 3
5 1 3 2 4 1 1 2 3 4

Sample Output 3

3

题意：有 n 个球，每个球上有一个序号 a[i]，现在要在球上重写序号，要求重写后，球上的新序号不同的个数不能大于 k，求最小的要重写的球的个数

思路：使用桶排简单统计即可

Source Program

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<string>
#include<cstring>
#include<cmath>
#include<ctime>
#include<algorithm>
#include<utility>
#include<stack>
#include<queue>
#include<vector>
#include<set>
#include<map>
#include<bitset>
#define EPS 1e-9
#define PI acos(-1.0)
#define INF 0x3f3f3f3f
#define LL long long
const int MOD = 1E9+7;
const int N = 200000+5;
const int dx[] = {-1,1,0,0,-1,-1,1,1};
const int dy[] = {0,0,-1,1,-1,1,-1,1};
using namespace std;

int a[N];
int bucket[N];
bool cmp(int a,int b){
    return a>b;
}
int main(){
    int n,k;
    scanf("%d%d",&n,&k);
    for(int i=1;i<=n;i++){
        scanf("%d",&a[i]);
        bucket[a[i]]++;
    }

    int cnt=0;
    for(int i=1;i<=n;i++){
        if(bucket[i]!=0){
            a[++cnt]=bucket[i];
        }
    }
    sort(a+1,a+1+cnt,cmp);
    int res=0;
    for(int i=k+1;i<=cnt;i++)
        res+=a[i];
    printf("%d\n",res);

    return 0;
}