Acdream 1427 Nice Sequence

Nice Sequence

Time Limit: 12000/6000MS (Java/Others)Memory Limit: 128000/64000KB (Java/Others)

Problem Description

Let us consider the sequence a1, a2,..., an of non-negative integer numbers. Denote as ci,j the number of occurrences of the number i among a1,a2,..., aj. We call the sequence k-nice if for all i1<i2 and for all j the following condition is satisfied: ci1,j ≥ ci2,j −k.

Given the sequence a1,a2,..., an and the number k, find its longest prefix that is k-nice.

Input

The first line of the input file contains n and k (1 ≤ n ≤ 200 000, 0 ≤ k ≤ 200 000). The second line contains n integer numbers ranging from 0 to n.

Output

Output the greatest l such that the sequence a1, a2,..., al is k-nice.

Sample Input

10 1
0 1 1 0 2 2 1 2 2 3
2 0
1 0

Sample Output

8
0

Source

Andrew Stankevich Contest 23

Manager

mathlover

解题：线段树。。。哎。。。当时不知怎么写复杂了。。。。真是蛋疼。。。。

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <cmath>
 5 #include <algorithm>
 6 #include <climits>
 7 #include <vector>
 8 #include <queue>
 9 #include <cstdlib>
10 #include <string>
11 #include <set>
12 #include <stack>
13 #define LL long long
14 #define pii pair<int,int>
15 #define INF 0x3f3f3f3f
16 using namespace std;
17 const int maxn = 210000;
18 struct node{
19     int lt,rt,minv,maxv;
20 };
21 node tree[maxn<<2];
22 void build(int lt,int rt,int v){
23     tree[v].lt = lt;
24     tree[v].rt = rt;
25     tree[v].minv = tree[v].maxv = 0;
26     if(lt == rt) return;
27     int mid = (lt+rt)>>1;
28     build(lt,mid,v<<1);
29     build(mid+1,rt,v<<1|1);
30 }
31 void update(int k,int v,int &val){
32     if(tree[v].lt == tree[v].rt){
33         val = tree[v].maxv = ++tree[v].minv;
34         return;
35     }
36     int mid = (tree[v].lt + tree[v].rt)>>1;
37     if(k <= mid) update(k,v<<1,val);
38     else if(k > mid) update(k,v<<1|1,val);
39     tree[v].minv = min(tree[v<<1].minv,tree[v<<1|1].minv);
40     tree[v].maxv = max(tree[v<<1].maxv,tree[v<<1|1].maxv);
41 }
42 int query_min(int lt,int rt,int v){
43     if(tree[v].lt == lt && tree[v].rt == rt) return tree[v].minv;
44     int mid = (tree[v].lt + tree[v].rt)>>1;
45     if(rt <= mid) return query_min(lt,rt,v<<1);
46     else if(lt > mid) return query_min(lt,rt,v<<1|1);
47     else return min(query_min(lt,mid,v<<1),query_min(mid+1,rt,v<<1|1));
48 }
49 int query_max(int lt,int rt,int v){
50     if(tree[v].lt == lt && tree[v].rt == rt) return tree[v].maxv;
51     int mid = (tree[v].lt + tree[v].rt)>>1;
52     if(rt <= mid) return query_max(lt,rt,v<<1);
53     else if(lt > mid) return query_max(lt,rt,v<<1|1);
54     else return max(query_max(lt,mid,v<<1),query_max(mid+1,rt,v<<1|1));
55 }
56 int n,k,val,tmp,ans;
57 int main() {
58     while(~scanf("%d %d",&n,&k)){
59         build(0,n + 1,1);
60         ans = 0;
61         bool flag = true;
62         for(int i = 1; i <= n; i++){
63             scanf("%d",&tmp);
64             update(tmp,1,val);
65             if(flag && val <= query_min(0,tmp,1)+k && val >= query_max(tmp+1,n+1,1)-k)
66                 ans = i;
67             else flag = false;
68         }
69         printf("%d\n",ans);
70     }
71     return 0;
72 }