Mishka received a gift of multicolored pencils for his birthday! Unfortunately he lives in a monochrome world, where everything is of the same color and only saturation differs. This pack can be represented as a sequence a1, a2, ..., an of n integer numbers — saturation of the color of each pencil. Now Mishka wants to put all the mess in the pack in order. He has an infinite number of empty boxes to do this. He would like to fill some boxes in such a way that:
- Each pencil belongs to exactly one box;
- Each non-empty box has at least k pencils in it;
- If pencils i and j belong to the same box, then |ai - aj| ≤ d, where |x| means absolute value of x. Note that the opposite is optional, there can be pencils i and j such that |ai - aj| ≤ d and they belong to different boxes.
Help Mishka to determine if it's possible to distribute all the pencils into boxes. Print "YES" if there exists such a distribution. Otherwise print "NO".
The first line contains three integer numbers n, k and d (1 ≤ k ≤ n ≤ 5·105, 0 ≤ d ≤ 109) — the number of pencils, minimal size of any non-empty box and maximal difference in saturation between any pair of pencils in the same box, respectively.
The second line contains n integer numbers a1, a2, ..., an (1 ≤ ai ≤ 109) — saturation of color of each pencil.
Print "YES" if it's possible to distribute all the pencils into boxes and satisfy all the conditions. Otherwise print "NO".
6 3 10 7 2 7 7 4 2
YES
6 2 3 4 5 3 13 4 10
YES
3 2 5 10 16 22
NO
In the first example it is possible to distribute pencils into 2 boxes with 3 pencils in each with any distribution. And you also can put all the pencils into the same box, difference of any pair in it won't exceed 10.
In the second example you can split pencils of saturations [4, 5, 3, 4] into 2 boxes of size 2 and put the remaining ones into another box.
题意：有n个铅笔，每只铅笔有个value，现在把他们分发到几个笔筒中，每个笔筒最少放k个笔，且放入的每支笔的value差不能大于d。求解能不能找到一种方案满足。
思路：
dp。先排序，判断最后一个数能否作为一个序列的结尾。
#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[500005];
bool dp[500005];
int main()
{
int n,k;
ll d;
scanf("%d%d%I64d",&n,&k,&d);
for(int i=1;i<=n;i++)
{
scanf("%I64d",&a[i]);
}
sort(a+1,a+1+n);
memset(dp,0,sizeof(dp));
dp[0]=1;
int p=1;
for(int i=0;i<=n;i++)
{
if(dp[i])//它是结尾的基础上，从下一个开始判断哪些可以作为结尾
{
p=max(p,i+k);//算过的就不用再重复算了，以防超时
//经典样例 50000 10 0
//1 1 1 1 1 1 1 1 1 1 1 ……
while(p<=n&&a[p]-a[i+1]<=d)
{//如果它可以作为结尾
dp[p]=1;
p++;
}
}
}
printf(dp[n]?"YES\n":"NO\n");
return 0;
}
