bzoj 4553 && HEOI2016 day1t3 seq

一个序列在所有变换中都单调不降的条件是i<j,a[i]<=min[j],mx[i]<=a[j],所以套CDQ就行了。

 #include<iostream>
 #include<cstdio>
 #include<algorithm>
 #include<cstring>
 #define N 100005
 #define inf 0x3f3f3f3f
 using namespace std;
 int n,m;
 int a[N];
 int f[N];
 int mx[N],mn[N];
 int tmp[N];
 int v[N];
 int res;
 bool cmp(int x,int y)
 {
     if(x<=res&&y<=res)return mx[x]<mx[y];
     if(x<=res)return mx[x]<=a[y];
     if(y<=res)return a[x]<mx[y];
     return a[x]<a[y];
 }
 int dp[N];
 struct node
 {
     int lazy,zhi;
 }aa[N*];
 void push_down(int x)
 {
     if(aa[x].lazy!=)
     {
         aa[x*].zhi=aa[x*+].zhi=-inf;
         aa[x*].lazy=aa[x*+].lazy=;
         aa[x].lazy=;
     }
     return ;
 }
 int qur(int x,int l,int r,int ll,int rr)
 {
     if(l>=ll&&r<=rr)
     {
         return aa[x].zhi;
     }
     push_down(x);
     int mid=(l+r)>>;
     if(rr<=mid)return qur(x*,l,mid,ll,rr);
     if(ll>mid)return qur(x*+,mid+,r,ll,rr);
     return max(qur(x*,l,mid,ll,rr),qur(x*+,mid+,r,ll,rr));
 }
 void gai(int x,int l,int r,int pos,int z)
 {
      if(l==r)
      {
          aa[x].zhi=max(aa[x].zhi,z);return ;
      }
      push_down(x);
      int mid=(l+r)>>;
      if(pos<=mid)gai(x*,l,mid,pos,z);
      else gai(x*+,mid+,r,pos,z);
      aa[x].zhi=max(aa[x*].zhi,aa[x*+].zhi);
 }
 void solve(int l,int r)
 {
     if (l==r)
     {
         f[l]=max(f[l],);
         return ;
     }
     int mid=(l+r)>>;
     solve(l,mid);res=mid;
     int cnt=;
     for(int i=l;i<=r;i++)tmp[++cnt]=i;
     sort(tmp+,tmp+cnt+,cmp);
     aa[].lazy=;aa[].zhi=-inf;
     for(int i=;i<=cnt;i++)
     {
         if(tmp[i]<=mid)gai(,,,a[tmp[i]],f[tmp[i]]);
         else f[tmp[i]]=max(f[tmp[i]],qur(,,,,mn[tmp[i]])+);
     }
     solve(mid+,r);
 }
 int main()
 {
    scanf("%d%d",&n,&m);
    memset(mx,0xcf,sizeof(mx));
    memset(mn,0x3f,sizeof(mn));
    for(int i=;i<=n;i++)
    {
           scanf("%d",&a[i]);
           mx[i]=mn[i]=a[i];
    }
    for(int i=;i<=m;i++)
    {
            int t1,t2;
            scanf("%d%d",&t1,&t2);
            mx[t1]=max(mx[t1],t2);
            mn[t1]=min(mn[t1],t2);
    }
    solve(,n);
    int ans=;
    for(int i=;i<=n;i++)ans=max(ans,f[i]);
    printf("%d\n",ans);
    return ;
 }