CodeForces 985 E Pencils and Boxes

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Pencils and Boxes

题意: n支铅笔,每只笔有一个颜色值, 无限个盒子, 如果一个盒子里面放笔的话, 就至少需要k支笔装在一起,并且一个盒子里面的笔的颜色值只差 不能大于 d。如果满足条件就输出YES,不满足就输出NO

题解:

对于这一条直线来说, B,C的长度为K+1, && val[C] - vla[A+1] <= d 如果我们说点C能够和前面的成员塞到一个盒子里面能成功的话, 那么[A,B]之间至少有一个点在前面处理过程中是可以分配的,设这个点为D, 那么[0,D], [D+1,C]都是合法的了。 

代码:

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 #define Fopen freopen("_in.txt","r",stdin); freopen("_out.txt","w",stdout);
 4 #define LL long long
 5 #define ULL unsigned LL
 6 #define fi first
 7 #define se second
 8 #define pb push_back
 9 #define lson l,m,rt<<1
10 #define rson m+1,r,rt<<1|1
11 #define max3(a,b,c) max(a,max(b,c))
12 #define min3(a,b,c) min(a,min(b,c))
13 typedef pair<int,int> pll;
14 const int INF = 0x3f3f3f3f;
15 const LL mod = 1e9+7;
16 const int N = 5e5+100;
17 int a[N];
18 int vis[N];
19 int main(){
20     ///Fopen;
21     int n, k, d;
22     scanf("%d%d%d", &n, &k, &d);
23     for(int i = 1; i <= n; i++) scanf("%d", &a[i]);
24     sort(a+1, a+1+n);
25     vis[0] = 1;
26     int pre  = 0, last = 0;
27     int cnt = 1;
28     for(int i = k; i <= n; i++){
29         while(pre <= last && a[i] - a[pre+1] > d){
30             if(vis[pre]) cnt--;
31             pre++;
32         }
33         if(cnt > 0) vis[i] = 1;
34         last++;
35         if(vis[last]) cnt++;
36     }
37     if(vis[n]) printf("YES\\n");
38     else printf("NO\\n");
39     return 0;
40 }
985E

 

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