cf Educational Codeforces Round 44 C. Liebig's Barrels

原题：
C. Liebig’s Barrels
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
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.

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 ≤ 10^5, 1 ≤ n·k ≤ 10^5, 0 ≤ l ≤ 10^9).

The second line contains m = n·k space-separated integers a1, a2, …, am (1 ≤ ai ≤ 10^9) — 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
input
4 2 1
2 2 1 2 3 2 2 3
output
7
input
2 1 0
10 10
output
20
input
1 2 1
5 2
output
2
inputCopy
3 2 1
1 2 3 4 5 6
output
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.

中文：
给你n×k个板子，每个板子长度为ai，现在让你把这n×k个板子分成n组，每组k个，每组当中取长度最小的值求和，且要求在这些组当中，任意取两组，两个组中各自的最小值相差不超过l，现在问你这个和最大是多少？

代码：

#include<iostream>
#include<limits>
#include<vector>
#include<string>
#include<sstream>
#include<algorithm>
using namespace std;
typedef long long ll;int n, k, l;
ll a[100001];int main()
{ios::sync_with_stdio(false);while (cin >> n >> k >> l){int ter = n*k;for (int i = 1; i <= ter; i++)cin >> a[i];sort(a + 1, a + 1 + ter);int ind = ter;for (int i = 2; i <= ter; i++){if (a[i] - a[1]>l){ind = i - 1;break;}}ll ans = 0;if (n == 1){cout << a[1] << endl;continue;}if (ind<n){cout << 0 << endl;continue;}if (k == 1){for (int i = 1; i <= ind; i++)ans += a[i];cout << ans << endl;continue;}int res = ter - ind, pre = ind, flag = 1;for (int i = ind; i >= 1;){if (res>0){if (res >= (k - 1)){res -= (k - 1);ans += a[i];i--;pre = i;}else{while ((pre - i + 1) + res<k)i--;ans += a[i];flag = 0;res = 0;}}else{if (flag != -1){if (flag!=0)i++;flag = -1;}i -= k;if (i<1)break;ans += a[i];}}cout << ans << endl;}return 0;
}

思路：

要求所有组中最小值的和最大，听起来像是二分，贪心可以解决。那就是要让每组当中的最小值从最大的值开始选取即可。

首先从小到大排序，找到最后一个小于最小值加上l的下标ind，从ind向前依次取值作为每一组当中的最小长度，同时判断在下标ind后面是否有元素，如果有，尽量选取k个，与前面选取到的最小长度组成一组。

如果ind后面没有元素了，那么就一次从后向前组成一组即可。

例如
4 2 2
1 2 2 3 3 4 5 6

上面序列是排好序的，最小值是1，加上l=2以后是3。

那么ind值等于5

从ind向前依次选取最小元素与ind后面的值组成一组
[3,4]
[3.5]
[2,6]

4,5,6都被选抽走后，依次从后往前组成组即可

[1,2]

结果为9

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