Flowers（priority_queue优先队列）(pta)

Recently Jack becomes much more romantic. He would like to prepare several bunches of flowers.

Each bunch of flowers must have exactly M flowers. As Jack does not want to be boring, he hopes that flowers in the same bunch are all different species. Now there are N species of flowers in the flower shop, and the number of the i-th species of flower is ai. Now Jack would like to know how many bunches of flowers he can prepare at most.

(Flowers are used to propose.)

Input

The first line contains an integer T (1≤T≤10) --- the number of test cases.

In the first line of each test case, there are two integers N, M (1≤N,M≤300000) --- the number of flowers' species and the number of flowers in a bunch.

In the second line of each test case, there are N integers --- the i-th integer indicates ai (1≤ai≤109), the number of i-th species' flowers.

Output

For each test case, output one integer in one line --- the answer of the corresponding test case.

Sample Input

1
5 3
1 1 1 2 1

Sample Output

题意：

花店有n种花，每种花有ai朵。

一个人要以束为单位买花，每束花有m朵，并且是m种不同类型的。

问他最多能买多少束。

思路：

来自身边大神的引导

优先队列是最合适此题的数据结构，可以出队、进队，还能自动排序（升序）。

每次取出剩余数量最多的m种花，把这m种花都减去其中最少的那种花的数量，也就是说又凑出了这么多束，答案进行累加。如果减完之后还有剩余，则重新加入队列。

需要注意的一点是，等剩余的花的种类不多的时候，这种方法将不再是最佳。那么就需要一束一束的减，相当于每凑出一束，又进行排序。

爱吃代码：

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<cmath>
#include<queue>
using namespace std;
priority_queue<int >q;
int s[300010];
int main()
{
    int t,i,n,m,a;
    long long ans;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&m);
        while(!q.empty())
            q.pop();
        for(i=0;i<n;i++)
        {
            scanf("%d",&a);
            q.push(a);
        }
        ans=0;
        while(!q.empty())
        {
            if(q.size()<=2*m)
                break;
            for(i=0;i<m;i++)
            {
                s[i]=q.top();
                q.pop();
            }
            for(i=0;i<m-1;i++)
            {
                if(s[i]-s[m-1]>0)
                    q.push(s[i]-s[m-1]);
                else
                    break;
            }
            ans+=s[m-1];
        }
        while(!q.empty())
        {
            if(q.size()<m)
                break;
            for(i=0;i<m;i++)
            {
                s[i]=q.top();
                q.pop();
            }
            for(i=0;i<m;i++)
            {
                if(s[i]-1>0)
                    q.push(s[i]-1);
                else
                    break;
            }
            ans++;
        }
        printf("%lld\n",ans);
    }
    return 0;
}

来源：CSDN

作者：涂涂817

链接：https://blog.csdn.net/weixin_43790882/article/details/104717341

标签

flowers

优先队列