【HDOJ6635】Nonsense Time（时间倒流，lis）

题意：给定n个数的数列，第i个数为a[i]，刚开始所有位置都处于禁用状态，第i次之后位置p[i]变为可用，求每次变化后的lis长度

n，a[i]，p[i]<=5e4

保证a[i]，p[i]均为随机生成的排列

思路：不知道非随机版本能不能树套树解决

 #include<bits/stdc++.h>
 using namespace std;
 typedef long long ll;
 typedef unsigned int uint;
 typedef unsigned long long ull;
 typedef pair<int,int> PII;
 typedef pair<ll,ll> Pll;
 typedef vector<int> VI;
 typedef vector<PII> VII;
 //typedef pair<ll,ll>P;
 #define N  100010
 #define M  200010
 #define fi first
 #define se second
 #define MP make_pair
 #define pb push_back
 #define pi acos(-1)
 #define mem(a,b) memset(a,b,sizeof(a))
 #define rep(i,a,b) for(int i=(int)a;i<=(int)b;i++)
 #define per(i,a,b) for(int i=(int)a;i>=(int)b;i--)
 #define lowbit(x) x&(-x)
 #define Rand (rand()*(1<<16)+rand())
 #define id(x) ((x)<=B?(x):m-n/(x)+1)
 #define ls p<<1
 #define rs p<<1|1

 const int MOD=1e9+,inv2=(MOD+)/;
       double eps=1e-;
       int INF=1e9;
       int inf=0x7fffffff;
       int dx[]={-,,,};
       int dy[]={,,-,};

 struct node
 {
     int x,y;
 }q[N];

 int dp[N],a[N],b[N],c[N],d[N],pre[N],ans[N],n;

 int read()
 {
    int v=,f=;
    char c=getchar();
    while(c<||<c) {if(c=='-') f=-; c=getchar();}
    while(<=c&&c<=) v=(v<<)+v+v+c-,c=getchar();
    return v*f;
 }

 int lis()
 {
     q[].x=;
     q[].y=;
     rep(i,,n)
     {
         q[i].x=INF;
         q[i].y=;
     }
     rep(i,,n) pre[i]=;
     int s=;
     rep(i,,n)
      if(c[i])
      {
         s++;
         int l=,r=s,last=;
         while(l<=r)
         {
             int mid=(l+r)>>;
             if(q[mid].x<=a[i]){last=mid; l=mid+;}
              else r=mid-;
         }
         dp[i]=last; pre[i]=q[last].y;
         if(q[last+].x>a[i]||q[last+].y==)
         {
             q[last+].x=a[i];
             q[last+].y=i;
         }
      }
     int res=,k=;
     rep(i,,n)
      if(c[i]&&dp[i]>res){res=dp[i]; k=i;}
     rep(i,,n) d[i]=;
     while(k)
     {
         d[k]=;
         k=pre[k];
     }
     return res;
 }

 void solve()
 {
     n=read();
     rep(i,,n) a[i]=read();
     rep(i,,n) b[i]=read();
     rep(i,,n) c[i]=;
     ans[n]=lis();
     per(i,n-,)
     {
         c[b[i+]]=;
         if(d[b[i+]]) ans[i]=lis();
          else ans[i]=ans[i+];
     }
     rep(i,,n-) printf("%d ",ans[i]);
     printf("%d\n",ans[n]);
 }

 int main()
 {
     int cas=read();
     while(cas--) solve();
     return ;
 }