Nice Sequence_线段树***

Description

      Let us consider the sequence a₁, a₂,..., a_n of non-negative integer numbers. Denote as c_i,j the number of occurrences of the number i among a₁,a₂,..., a_j. We call the sequence k-nice if for all i₁<i₂ and for all j the following condition is satisfied: c_i1,j ≥ c_i2,j −k. 

      Given the sequence a₁,a₂,..., a_n 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 a ₁, a ₂,..., a _l is k-nice.

Sample Input

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

Sample Output

8
1

【题意】用线段树维护 0到A[i]-1间的最小值，用F[A[i]] 统计频率。判断 0 到 A[i]-1范围内的最小值与F[A[i]]-K的大小即可。

【思路】cnt[i]是当前i出现的次数。每次读入一个数a进来，cnt[a]++。只要c[1]...c[a-1]都满足c[i]+K>=c[a]就可以了，一旦不满足就结束程序并输出答案。

那么找c[1]...c[a-1]中最小的即可。如果最小的都满足不等式就不用验证其他的了。于是用线段树实现这个查询操作。

注意线段树建树从[0,N]开始，因为The second line contains n integer numbers ranging from 0 to n.

参考：http://blog.csdn.net/liao_jingyi/article/details/43531123

#include<iostream>
#include<stdio.h>
#include<string.h>
using namespace std;
const int N=200000+10;
int cnt[N];
int n,k;
struct node
{
    int l,r,sum;
} tree[N*4];
void build(int k,int l,int r)
{
    tree[k].l=l;
    tree[k].r=r;
    if(l==r) return ;
    int mid=l+r>>1;
    build(k*2,l,mid);
    build(k*2+1,mid+1,r);
}
void update(int x,int k)
{
    if(tree[k].l==tree[k].r)
    {
        tree[k].sum++;
        return ;
    }
    int mid=(tree[k].l+tree[k].r)>>1;
    if(x<=mid) update(x,k*2);
    else update(x,k*2+1);
    tree[k].sum=min(tree[k*2].sum,tree[k*2+1].sum);
}

int query(int x,int k)
{
    if(x>=tree[k].r) return tree[k].sum;
    int mid=(tree[k].l+tree[k].r)>>1;
    int ansl=query(x,k*2);
    int ansr=0x3f3f3f3f;
    if(mid<x) ansr=query(x,k*2+1);
    return min(ansl,ansr);
}

int main()
{
    scanf("%d%d",&n,&k);
    build(1,0,n);
    int ans=0;
    for(int i=1; i<=n; i++)
    {
        int a;
        scanf("%d",&a);
        cnt[a]++;
        update(a,1);
        if(query(a,1)+k>=cnt[a])
            ans=i;
        else break;
    }
    printf("%d
",ans);

    return 0;
}