[bzoj4491]我也不知道题目名字是什么

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给定一个序列A[i]，每次询问l,r，求[l,r]内最长子串，使得该子串为不上升子串或不下降子串

n,q<=50000

线段树维护区间左右是谁，从左端开始／到右端结束的最长上升／下降子串长度即可。

复杂度nlogn

#include<iostream>
#include<cstdio>
#define MN 50000
using namespace std;
inline int read()
{
    int x = 0 , f = 1; char ch = getchar();
    while(ch < '0' || ch > '9'){ if(ch == '-') f = -1;  ch = getchar();}
    while(ch >= '0' && ch <= '9'){x = x * 10 + ch - '0';ch = getchar();}
    return x * f;
}

struct data{int l1,l2,r1,r2,ans,L,R,len;
    data(){}
    data(int x){l1=l2=r1=r2=ans=len=1;L=R=x;}
    friend data operator +(const data&a,const data&b)
    {
        data c;c.L=a.L;c.R=b.R;c.len=a.len+b.len;
        c.ans=max(max(a.ans,b.ans),max(a.r2,b.l2));
        if(b.L>=a.R) c.ans=max(c.ans,a.r1+b.l1);
        if(b.L<=a.R) c.ans=max(c.ans,a.r2+b.l2);
        c.l1=a.l1==a.len?a.l1+(b.L>=a.R?b.l1:0):a.l1;
        c.l2=a.l2==a.len?a.l2+(b.L<=a.R?b.l2:0):a.l2;
        c.r1=b.r1==b.len?b.r1+(b.L>=a.R?a.r1:0):b.r1;
        c.r2=b.r2==b.len?b.r2+(b.L<=a.R?a.r2:0):b.r2;
        return c;
    }
};
struct Tree{int l,r;data x;}T[MN*4+5];
int n,m,a[MN+5];

void build(int x,int l,int r)
{
    if((T[x].l=l)==(T[x].r=r)){T[x].x=data(a[l]);return;}
    int mid=l+r>>1;
    build(x<<1,l,mid);build(x<<1|1,mid+1,r);
    T[x].x=T[x<<1].x+T[x<<1|1].x;
}

data Query(int x,int l,int r)
{
    if(T[x].l==l&&T[x].r==r) return T[x].x;
    int mid=T[x].l+T[x].r>>1;
    if(r<=mid) return Query(x<<1,l,r);
    else if(l>mid) return Query(x<<1|1,l,r);
    else return Query(x<<1,l,mid)+Query(x<<1|1,mid+1,r);
}

int main()
{
    n=read();
    for(int i=1;i<=n;++i)a[i]=read();
    m=read();build(1,1,n);
    for(int i=1;i<=m;++i)
    {
        int l=read(),r=read();
        data x=Query(1,l,r);
        x.ans=max(x.ans,max(x.l1,x.l2));
        x.ans=max(x.ans,max(x.r1,x.r2));
        printf("%d
",x.ans);
    }
    return 0;
}