Liebig‘s Barrels
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard 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?≤?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
input
Copy
4 2 12 2 1 2 3 2 2 3
output
Copy
7
input
Copy
2 1 010 10
output
Copy
20
input
Copy
1 2 15 2
output
Copy
2
input
Copy
3 2 11 2 3 4 5 6
output
Copy
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：输入 n k l 你要做n个桶，每个桶需要k个木板，用木板拼好的桶相互之间体积的差距<=l，桶的体积大小就是最短的那根木板的长度大小。
第二行 共n*k个数，分别表示n*k个木板的长度。
分析：
先对边排个序
不存在的情况，就是a[n]-a[1]>l，那就是不存在，因为要是差距尽可能小，前n小的都分别作为n个桶的一块木板，那么这之中最大的差距就是a[n]-a[1]，要是a[n]-a[1]都满足条件（<=l）了，那就满足条件了。
其次，要使体积和最大输出体积和，我毛想想觉得s=a[1]+……a[n]，结果WA了，引起了我的深思。
因为：
eg：4 3 17
1 2 3 5 9 13 18 21 22 23 25 26
它可以这样组3组：
18 25 26
13 22 23
1 2 3
5 9 21
这样体积为1+5+13+18=37，不是简单地1 +2 +3 +5=11
所以我的思路：先要找到最大的满足条件的数，可以用二分找更快，在这组样例中，是18，它-a[1]<=l，
那么从最后开始去k-1个和18拼，s+=18，再下一个数13（25 26），再从最后找k-1个数（22 23），
再下一个数9，发现再k-1个数不够了，那就从头开始找了，（1 2 3）一组，在去（5 9 13）时，发现13
已经被取走，那就s+=5就可以了。
#include <iostream>
#include<cstring>
#include<string>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<deque>
#include<vector>
#define ll long long
#define inf 0x3f3f3f3f
#define mod 1000000007;
using namespace std;
ll a[100005];
bool cmp(ll a,ll b)
{
return a<b;
}
int main()
{
ll n,k,l;
scanf("%I64d%I64d%I64d",&n,&k,&l);
for(ll i=1;i<=n*k;i++)
{
scanf("%I64d",&a[i]);
}
sort(a+1,a+1+n*k,cmp);
if(a[n]-a[1]>l)
{
printf("0");
}
else
{
ll s=0;
ll p=-1;
for(ll i=n*k;i>=1;i--)
{
if(a[i]-a[1]<=l)
{
p=i;//找到标准数，最大的满足条件的数
break;
}
}
s=0;
int num=0;//记录从标准数向前取了多少
int j=p;
for(ll i=n*k;i-(k-1)>p;i=i-(k-1))//先从后往前取
{
s+=a[j--];
num++;
}
for(ll i=1;i<p-num+1;i=i+k)//在从前往后取
{
s+=a[i];
}
printf("%I64d",s);
}
return 0;
}
codeforce 985C Liebig's Barrels（贪心+思维）
原文地址：https://www.cnblogs.com/caiyishuai/p/9080572.html