BZOJ 3904 longest ascending subsequence lkids line segment tree

BZOJ 3904 longest ascending subsequence lkids line segment tree

Given a sequence, evaluate the sub-series of sertices starting with a decimal number so that the difference between the two adjacent items is not less than k

F [I] [0] indicates that the number of I is the largest and largest of the sequences.

F [I] [1] indicates that the number of I is the largest of the smaller values in the sequence.

The violent transfer is from O (n ^ 2 ).

We found that the values of decision points are continuous intervals, so we can use the line segment tree to maintain them.

(Really simple)

#include 
 
  #include 
  
   #include 
   
    #include #define M 200200using namespace std;template
    
      struct Segtree{Segtree *ls,*rs;int val;Segtree():ls(0x0),rs(0x0),val(_) {}friend void Update(Segtree *&p,int x,int y,int l,int r,int val){int mid=x+y>>1;if(!p) p=new Segtree;if(x==l&&y==r){p->val=max(p->val,val);return ;}if(r<=mid)Update(p->ls,x,mid,l,r,val);else if(l>mid)Update(p->rs,mid+1,y,l,r,val);elseUpdate(p->ls,x,mid,l,mid,val) , Update(p->rs,mid+1,y,mid+1,r,val);}friend int Get_Ans(Segtree *p,int x,int y,int pos){int mid=x+y>>1;if(!p) return _;if(x==y) return p->val;if(pos<=mid)return max(Get_Ans(p->ls,x,mid,pos),p->val);elsereturn max(Get_Ans(p->rs,mid+1,y,pos),p->val);}};Segtree< 0> *tree0=new Segtree< 0>;Segtree<-1> *tree1=new Segtree<-1>;int n,k,ans;int a[M],f[M][2];int main(){int i;cin>>n>>k;for(i=1;i<=n;i++){scanf("%d",&a[i]);if(a[i]-k>=0){f[i][0]=Get_Ans(tree1,0,100000000,a[i]-k)+1;Update(tree0,0,100000000,0,a[i],f[i][0]);}if(a[i]+k<100000000){f[i][1]=Get_Ans(tree0,0,100000000,a[i]+k)+1;Update(tree1,0,100000000,a[i],100000000,f[i][1]);}ans=max(ans,f[i][0]);ans=max(ans,f[i][1]);}cout<