• LCIS


    LCIS
    Time Limit: 3000 MS Memory Limit: 32768 K
    Total Submit: 70(28 users) Total Accepted: 33(26 users) Rating: Special Judge: No
    Description
    Given n integers.
    You have two operations:
    U A B: replace the Ath number by B. (index counting from 0)
    Q A B: output the length of the longest consecutive increasing subsequence (LCIS) in [a, b].
    Input
    T in the first line, indicating the case number.
    Each case starts with two integers n , m(0<n,m<=105).
    The next line has n integers(0<=val<=105).
    The next m lines each has an operation:
    U A B(0<=A,n , 0<=B=105)
    OR
    Q A B(0<=A<=B< n).
    Output
    For each Q, output the answer.
    Sample Input
    1 10 10 7 7 3 3 5 9 9 8 1 8 Q 6 6 U 3 4 Q 0 1 Q 0 5 Q 4 7 Q 3 5 Q 0 2 Q 4 6 U 6 10 Q 0 9
    Sample Output
    1 1 4 2 3 1 2 5
    Recommend
    `Wind @Hrbust
    Author
    shǎ崽
    #include<cstdio>
    #include<cmath>
    #include<queue>
    const int SIZE=100005;
    int LC[SIZE<<2];
    int RC[SIZE<<2];
    int CS[SIZE<<2];
    int a[SIZE];
    int N,M;
    template <class typeone>
    typeone max(typeone x,typeone y)
    {
        return x>y? x:y;
    }
    template <class typeone >
    typeone min(typeone x,typeone y)
    {
        return x<y? x:y;
    }
    void pushup(int l,int r,int lr)
    {
        LC[lr]=LC[lr<<1];
        RC[lr]=RC[lr<<1|1];
        CS[lr]=0;
        int m=(l+r)>>1;
        if(a[m]<a[m+1])
        {
            CS[lr]=RC[lr<<1]+LC[lr<<1|1];
            if(LC[lr]==m-l+1)
            {
                LC[lr]+=LC[lr<<1|1];
            }
            if(RC[lr]==r-m)
            {
                RC[lr]+=RC[lr<<1];
            }
        }
        CS[lr]=max(CS[lr],max(CS[lr<<1],CS[lr<<1|1]));
        return;
    }
    
    void build(int l,int r,int rt)
    {
        LC[rt]=RC[rt]=CS[rt]=0;
        if(l==r)
        {
            LC[rt]=RC[rt]=CS[rt]=1;
            scanf("%d",&a[l]);
            return ;
        }
        int m=(l+r)>>1;
        build(l,m,rt<<1);
        build(m+1,r,rt<<1|1);
        pushup(l,r,rt);
        return ;
    }
    void update(int pos,int l,int r,int rt,int val)
    {
        if(l==pos&&r==pos)
        {
            a[pos]=val;
            return;
        }
        int m=(l+r)>>1;
        if(pos<=m)
        {
            update(pos,l,m,rt<<1,val);
        }
        else
        {
            update(pos,m+1,r,rt<<1|1,val);
        }
        pushup(l,r,rt);
        return ;
    }
    int query(int L,int R,int l,int r,int rt)
    {
        if(L<=l&&R>=r)
        {
            return CS[rt];
        }
        int m=(l+r)>>1;
        if(R<=m)
        {
            return query(L,R,l,m,rt<<1);
        }
        else if(L>m)
        {
            return  query(L,R,m+1,r,rt<<1|1);
        }
        else
        {
            int t1=0,t2=0,t3=0;
            t1=query(L,R,l,m,rt<<1);
            t2=query(L,R,m+1,r,rt<<1|1);
            if(a[m]<a[m+1])
            {
                t3=min(LC[rt<<1|1],R-m)+min(RC[rt<<1],m-L+1);
            }
            return max(t3,max(t1,t2));
        }
    }
    int main()
    {
        int T;
        while(~scanf("%d",&T))
        {
            while(T--)
            {
                char k[2];
                scanf("%d %d",&N,&M);
                build(1,N,1);
                int x,y;
            //    getchar();
                while(M--)
                {
                    scanf("%s %d %d",&k,&x,&y);
                    if(strcmp(k,"Q")==0)
                    {
                        printf("%d
    ",query(x+1,y+1,1,N,1));    
                    }
                    else
                    {
                        update(x+1,1,N,1,y);
                    }
                }
            }
        }
        return 0;
    }
  • 相关阅读:
    Android TextView中的ellipsize属性
    Android 仿微信点赞和评论弹出框
    Java 数组倒序
    Android List去掉重复数据
    Android HttpClient get传递数组
    android HttpClient
    kali中的APT软件包处理工具(apt-get)、Debian软件包管理器(dpkg)、源代码压缩和Nessus安装实用指南
    安装Kali里的应用程序或软件包
    渗透测试方法论
    Kali Linux的介绍
  • 原文地址:https://www.cnblogs.com/beige1315402725/p/4931906.html
Copyright © 2020-2023  润新知