Codeforces 985C 贪心

最新推荐文章于 2021-12-31 18:38:01 发布

原创最新推荐文章于 2021-12-31 18:38:01 发布 · 427 阅读

0 ·

CC 4.0 BY-SA版权

本文介绍了一种算法，用于解决由n*k块木板拼接成n个木桶的问题，要求每个木桶使用k块木板且任意两桶间的容量差不超过l，目标是求得所有木桶的最大总容量。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

You have m = n·k wooden staves. The i-th stave has length ai. You have to assemble n barrels consisting of k staves each, you can use any k staves to construct a barrel. Each stave must belong to exactly one barrel.

Let volume vj of barrel j be equal to the length of the minimal stave in it.

$\text{[math]}$

You want to assemble exactly n barrels with the maximal total sum of volumes. But you have to make them equal enough, so a difference between volumes of any pair of the resulting barrels must not exceed l, i.e. |vx - vy| ≤ l for any 1 ≤ x ≤ n and 1 ≤ y ≤ n.

Print maximal total sum of volumes of equal enough barrels or 0 if it's impossible to satisfy the condition above.

Input

The first line contains three space-separated integers n, k and l (1 ≤ n, k ≤ 105, 1 ≤ n·k ≤ 105, 0 ≤ l ≤ 109).

The second line contains m = n·k space-separated integers a1, a2, ..., am (1 ≤ ai ≤ 109) — lengths of staves.

Output

Print single integer — maximal total sum of the volumes of barrels or 0 if it's impossible to construct exactly n barrels satisfying the condition |vx - vy| ≤ l for any 1 ≤ x ≤ n and 1 ≤ y ≤ n.

Examples
inputCopy
4 2 1
2 2 1 2 3 2 2 3
outputCopy
7
inputCopy
2 1 0
10 10
outputCopy
20
inputCopy
1 2 1
5 2
outputCopy
2
inputCopy
3 2 1
1 2 3 4 5 6
outputCopy
0

Note

In the first example you can form the following barrels: [1, 2], [2, 2], [2, 3], [2, 3].

In the second example you can form the following barrels: [10], [10].

In the third example you can form the following barrels: [2, 5].

In the fourth example difference between volumes of barrels in any partition is at least 2 so it is impossible to make barrels equal enough.

题意：木桶容量定义为最短的木板的长度，，用n*k块木板拼成n个木桶，要求每个木桶用k块木板，任意两个木桶容量差距小于l，求能拼成的木桶的最大容积和

思路：贪心，排序之后找到离最短木板差距小于等于l的最长木板，然后尽可能选择长的木板，具体过程见代码。

#include <bits/stdc++.h>

using namespace std;

typedef long long ll;
ll a[100005];

int main()
{
    ll n,k,l;
    scanf("%I64d%I64d%I64d",&n,&k,&l);
    for(int i=0; i<n*k; i++){
        scanf("%I64d",&a[i]);
    }
    sort(a,a+n*k);
    if(a[n-1]-a[0]>l){ //无法拼出n个桶
        cout<<0<<endl;
        return 0;
    }
    ll tail = lower_bound(a,a+n*k,a[0]+1+l)-a, pos = 0;     //tail作为可以作为最短木板的范围,pos为当前木板的位置
    ll cnt = tail-n;
    ll res = 0;
    for(int i=0; i<n; i++){
        res += a[pos];
        if(cnt >= k-1){    //若仍有大于k-1块木板,则用这k-1块木板拼成桶
            cnt -= k-1;
            pos += k;
        }
        else if(cnt){
            pos += cnt+1;
            cnt = 0;
        }
        else
            pos++;      //每次往后取一块木板
    }
    printf("%I64d\n",res);
    return 0;
}