• 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



    题解及代码:

          比赛的时候是队友写的这道题,由于查询的时候要进行优化,所以给他加了一个线段树维护区间最小值就AC了。

          今天看了一下这道题,感觉也不难。它的定义有些难懂啊。事实上就是当我们从左向右输入到一个元素x时,要保证其左側元素中1-(x-1)出现的次数大于等于 x的出现次数-k,找出这种最长的前缀。


    #include <iostream>
    #include <cstdio>
    #include <cstdlib>
    #include <cstring>
    #include <set>
    #include <map>
    #include <queue>
    #include <string>
    #define maxn 200010
    #define lson l,m,rt<<1
    #define rson m+1,r,rt<<1|1
    using namespace std;
    typedef long long  ll;
    int a[200005];
    
    struct segment
    {
        int l,r;
        int value;
    } son[maxn<<2];
    
    
    void PushUp(int rt)
    {
        son[rt].value=min(son[rt<<1].value,son[rt<<1|1].value);
    }
    
    
    void Build(int l,int r,int rt)
    {
        son[rt].l=l;
        son[rt].r=r;
        if(l==r)
        {
            son[rt].value=0;
            return;
        }
        int m=(l+r)/2;
        Build(lson);
        Build(rson);
        PushUp(rt);
    }
    
    
    void Update(int p,int rt)
    {
        if(son[rt].l==son[rt].r)
        {
            son[rt].value++;
            return;
        }
    
        int m=(son[rt].l+son[rt].r)/2;
        if(p<=m)
            Update(p,rt<<1);
        else
            Update(p,rt<<1|1);
    
        PushUp(rt);
    }
    
    
    int  Query(int l,int r,int rt)
    {
        if(son[rt].l==l&&son[rt].r==r)
        {
            return son[rt].value;
        }
    
        int ret=0;
        int m=(son[rt].l+son[rt].r)/2;
    
        if(r<=m)
            ret=Query(l,r,rt<<1);
        else if(l>m)
            ret=Query(l,r,rt<<1|1);
        else
        {
            ret=Query(lson);
            ret=min(ret,Query(rson));
        }
        return ret;
    }
    
    
    
    int main()
    {
        int n,k,x,ans=0;
        bool flag=false;
        scanf("%d%d",&n,&k);
        Build(1,n+1,1);
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&x);
            if(!flag)
            {
                x++;
                a[x]++;
                Update(x,1);
                if(Query(1,x,1)<a[x]-k)
                flag=true;
    
                if(!flag) ans=i;
            }
        }
        printf("%d
    ",ans);
        return 0;
    }
    






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  • 原文地址:https://www.cnblogs.com/liguangsunls/p/6691891.html
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