[BZOJ]4491: 我也不知道题目名字是什么 线段树（差分）

Description

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

这道题目有两个思路：1、对原序列差分，转化为经典问题。2、线段树维护6个东西，分别是左边开始的最长不上升序列、最长不下降序列长度，右边开始的最长不上升序列、最长不下降序列长度，本段内最长不上升序列、最长不下降序列长度。我写了后面的。代码：

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>using namespace std;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<<3)+(x<<1)+ch-'0';ch=getchar();}return x*f;}const int maxn=50050;int n,a[maxn];int trlen=0;struct Tree{int l,r,lc,rc,lup,ldown,rup,rdown,up,down;}tr[maxn*4];Tree pushup(Tree lc,Tree rc,int L,int R,int l,int r){Tree t;t.lc=L;t.rc=R;t.l=l;t.r=r;if(lc.lup==lc.r-lc.l+1&&lc.r<=n&&a[lc.r]<=a[lc.r+1])t.lup=lc.lup+rc.lup;else t.lup=lc.lup;if(lc.ldown==lc.r-lc.l+1&&lc.r<=n&&a[lc.r]>=a[lc.r+1])t.ldown=lc.ldown+rc.ldown;else t.ldown=lc.ldown;if(rc.rup==rc.r-rc.l+1&&rc.l>1&&a[rc.l-1]>=a[rc.l])t.rup=rc.rup+lc.rup;else t.rup=rc.rup;if(rc.rdown==rc.r-rc.l+1&&rc.l>1&&a[rc.l-1]<=a[rc.l])t.rdown=rc.rdown+lc.rdown;else t.rdown=rc.rdown;t.up=max(lc.up,rc.up);if(lc.r<=n&&a[lc.r]<=a[lc.r+1])t.up=max(t.up,lc.rdown+rc.lup);t.down=max(lc.down,rc.down);if(lc.r<=n&&a[lc.r]>=a[lc.r+1])t.down=max(t.down,lc.rup+rc.ldown);return t;}void build(int l,int r){int t=++trlen;tr[t].l=l;tr[t].r=r;if(l==r)tr[t].lup=tr[t].ldown=tr[t].rup=tr[t].rdown=tr[t].up=tr[t].down=1;if(l<r){int mid=l+r>>1;int lc=trlen+1;build(l,mid);int rc=trlen+1;build(mid+1,r);tr[t]=pushup(tr[lc],tr[rc],lc,rc,l,r);}}Tree query(int now,int l,int r){if(tr[now].l==l&&tr[now].r==r)return tr[now];int mid=tr[now].l+tr[now].r>>1,lc=tr[now].lc,rc=tr[now].rc;if(r<=mid)return query(lc,l,r);else if(l>mid)return query(rc,l,r);else return pushup(query(lc,l,mid),query(rc,mid+1,r),tr[now].lc,tr[now].rc,tr[now].l,tr[now].r);}int main(){n=read();for(int i=1;i<=n;i++)a[i]=read();build(1,n);int q=read();while(q--){int l=read(),r=read();Tree ans=query(1,l,r);printf("%d\n",max(ans.up,ans.down));}}