Codeforces 985 E

E - Pencils and Boxes

思路：

dp

先排个序，放进一个袋子里的显然是一段区间

定义状态：pos[i]表示小于等于i的可以作为（放进一个袋子里的）一段区间起点的离i最近的位置

显然，初始状态：pos[i] = 1,1 <= i <= k

状态转移：

pos[i+1] = i+1 ，如果 a[i] - a[pos[i-k+1]] <= d , 因为在这种情况下pos[i-k+1] 到 i 可以放进一个袋子里，那么i+1就可以作为新的起点了

pos[i+1] = pos[i], 其他情况

最后只要判断pos[n+1] 等不等于 n+1 就可以判断是不是都能放进袋子里

代码：

#include<bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define pi acos(-1.0) 
#define LL long long 
//#define mp make_pair
#define pb push_back
#define ls rt<<1, l, m
#define rs rt<<1|1, m+1, r
#define ULL unsigned LL
#define pll pair<LL, LL>
#define pii pair<int, int> 
#define mem(a, b) memset(a, b, sizeof(a))
#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define fopen freopen("in.txt", "r", stdin);freopen("out.txt", "w", stout);
//head

const int N = 5e5 + 5;
int a[N], pos[N];
int main() {
    int n, k, d;
    scanf("%d%d%d", &n, &k, &d);
    for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
    sort(a+1, a+1+n);
    for (int i = 1; i <= k; i++) pos[i] = 1;
    for (int i = k; i <= n; i++) {
        int pre = pos[i-k+1];
        if(a[i] - a[pre] <= d) pos[i+1] = i+1;
        else pos[i+1] = pos[i];
        //cout << i+1 << " " << pos[i+1] << endl; 
    }
    if(pos[n+1] == n+1) printf("YES
");
    else printf("NO
");
    return 0;
}