【ZOJ3349】Special Subsequence

题解：
线段树优化DP
令f[i]表示到 i 位置满足题意的子序列长度的最大值
转移显然 f[i]=max(f[j])+1 (abs(i-j)<=k)
所以只需要离散化后建一棵权值线段树，单点修改，区间查询最大值即可
当然你想写平衡树也可以啊

//by sdfzchy#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>#define lson L,mid,rt<<1#define rson mid+1,R,rt<<1|1using namespace std;typedef long long LL;const int inf=(1<<30),N=200100;int n,m;inline int in(){    char ch=getchar();    int f=1,tmp=0;    while(ch<'0'||ch>'9') {if(ch=='-') f=-1;ch=getchar();}    while(ch>='0'&&ch<='9') {tmp=(tmp<<1)+(tmp<<3)+(ch-'0');ch=getchar();}    return tmp*f;}int a[N],ha[N*4],f[N];int ma[N*12+5];void upd(int rt){    ma[rt]=max(ma[rt<<1],ma[rt<<1|1]);}void change(int pos,int x,int L,int R,int rt){    if(L==R) {ma[rt]=max(x,ma[rt]);return;}    int mid=(L+R)>>1;    if(pos<=mid) change(pos,x,lson);    else change(pos,x,rson);    upd(rt);}int query(int ll,int rr,int L,int R,int rt){    if(ll<=L&&rr>=R) return ma[rt];    int mid=(L+R)>>1,ret=0;    if(ll<=mid) ret=max(ret,query(ll,rr,lson));    if(rr>mid)  ret=max(ret,query(ll,rr,rson));    return ret;}int d;int l[N],r[N],ans;int main(){    while(~scanf("%d%d",&n,&d))    {        memset(f,0,sizeof(f));        memset(ma,0,sizeof(ma));        for(int i=1;i<=n;i++) a[i]=in(),ha[3*i-2]=a[i]-d,ha[3*i-1]=a[i],ha[3*i]=a[i]+d;        sort(ha+1,ha+3*n+1);        int len=unique(ha+1,ha+3*n+1)-ha-1;        for(int i=1;i<=n;i++)        {            l[i]=lower_bound(ha+1,ha+len+1,a[i]-d)-ha;            r[i]=lower_bound(ha+1,ha+len+1,a[i]+d)-ha;            a[i]=lower_bound(ha+1,ha+len+1,a[i])-ha;        }        for(int i=1;i<=n;i++)        {            f[i]=query(l[i],r[i],1,len,1)+1;            change(a[i],f[i],1,len,1);            ans=max(ans,f[i]);        }        printf("%d\n",ans);        ans=0;    }    return 0;}