You have a Petri dish with bacteria and you are preparing to dive into the harsh micro-world. But, unfortunately, you don't have any microscope nearby, so you can't watch them.
You know that you have nn bacteria in the Petri dish and size of the ii-th bacteria is aiai. Also you know intergalactic positive integer constant KK.
The ii-th bacteria can swallow the jj-th bacteria if and only if ai>ajai>aj and ai≤aj+Kai≤aj+K. The jj-th bacteria disappear, but the ii-th bacteria doesn't change its size. The bacteria can perform multiple swallows. On each swallow operation any bacteria ii can swallow any bacteria jjif ai>ajai>aj and ai≤aj+Kai≤aj+K. The swallow operations go one after another.
For example, the sequence of bacteria sizes a=[101,53,42,102,101,55,54]a=[101,53,42,102,101,55,54] and K=1K=1. The one of possible sequences of swallows is: [101,53,42,102,101––––,55,54][101,53,42,102,101_,55,54] →→ [101,53–––,42,102,55,54][101,53_,42,102,55,54] →→ [101––––,42,102,55,54][101_,42,102,55,54] →→ [42,102,55,54–––][42,102,55,54_] →→ [42,102,55][42,102,55]. In total there are 33 bacteria remained in the Petri dish.
Since you don't have a microscope, you can only guess, what the minimal possible number of bacteria can remain in your Petri dish when you finally will find any microscope.
The first line contains two space separated positive integers nn and KK (1≤n≤2⋅1051≤n≤2⋅105, 1≤K≤1061≤K≤106) — number of bacteria and intergalactic constant KK.
The second line contains nn space separated integers a1,a2,…,ana1,a2,…,an (1≤ai≤1061≤ai≤106) — sizes of bacteria you have.
Print the only integer — minimal possible number of bacteria can remain.
7 1 101 53 42 102 101 55 54
3
6 5 20 15 10 15 20 25
1
7 1000000 1 1 1 1 1 1 1
7
The first example is clarified in the problem statement.
In the second example an optimal possible sequence of swallows is: [20,15,10,15,20–––,25][20,15,10,15,20_,25] →→ [20,15,10,15–––,25][20,15,10,15_,25] →→ [20,15,10–––,25][20,15,10_,25] →→[20,15–––,25][20,15_,25] →→ [20–––,25][20_,25] →→ [25][25].
In the third example no bacteria can swallow any other bacteria.
题意:对任意一个数a[i] , 可以对任意 满足 i != j 且 a[i] > a[j] && a[i] <= a[j] +k 的 a[j] 可以被删掉,求使最终剩下的个数最少。
思路: 显然需要对数组排序,之后为了达到剩余个数最少,我们需要从小到大按照题意模拟扫一遍。
#include <bits/stdc++.h>
using namespace std;
const int maxn=1e6+10;
int n,a[maxn],k;
bool vis[maxn];
int main(){
memset(vis,false,sizeof(vis));
scanf("%d%d",&n,&k);
for (int i=1; i<=n ;i++) scanf("%d",&a[i]);
sort(a+1,a+1+n);
int cnt=0;
int i=1,j=1;
while(i<=n && j<=n){
if(a[i]>a[j] && a[i]<=a[j]+k){
vis[j]=1;
j++;
}
else if(a[i]<=a[j])i++;
else j++;
}
for (int i=1; i<=n ;i++) if(!vis[i])cnt++;
printf("%d
",cnt);
return 0;
}