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 ≤ 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：给你一些木板，k个木板可以组成一个木桶，体积为木桶最短的那个木板的长度，每个桶的体积不能相差超过l，问你可以组成的最大体积和为多少，如果不能组成输出0
思路：先小到大排序，然后找出小于第一个加l 的有多少个，因为第一个必要用到，所以贪心只能在满足<=a[0]+l的木板里进行。
接下来，直接扩展木板就可以了，就是让最短的木板尽可能的大
#include<bits/stdc++.h>
#define LL long long
#define Max 100005
const LL mod=1e9+7;
const LL LL_MAX=9223372036854775807;
using namespace std;
LL n,k,l;
LL a[Max];
priority_queue<LL ,vector<LL> ,greater<LL> >p;
int main()
{
scanf("%lld%lld%lld",&n,&k,&l);
LL x;
for(int i=0;i<n*k;i++){
scanf("%lld",&a[i]);
p.push(a[i]);
}
sort(a,a+(n*k));
LL ans=0,mn=p.top();
int t=0;
while(p.size()){
if(p.top()>(mn+l))//找有多少个小于a[0]+l的木板
break;
t++;
p.pop();
}
if(t<n){//如果小于a[0]+l的不足n个，很明显不满足条件
printf("0\n");
return 0;
}
int now = 0;
for(int i=0;i<n;i++){
ans+=a[now++];
for(int j=0;j<k-1;j++){//扩张木板
if(t-now > n-i-1)
now++;
else
break;
}
}
printf("%lld\n",ans);
return 0;
}