Acdream 1427 nice sequence

Nice sequencetime limit: 12000/6000 ms (Java/others) memory limit: 128000/64000 KB (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 LSuch that the sequence A1, A2,..., Al is k-nice.Sample Input

10 10 1 1 0 2 2 1 2 2 32 01 0

Sample output

80

Sourceandrew stankevich contest 23 managermathlover problem solving: Line Segment tree... Ah... At that time, I did not know how to write complicated .... It really hurts ....

 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 long14 #define pii pair<int,int>15 #define INF 0x3f3f3f3f16 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 }

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Acdream 1427 nice sequence