ZOJ-3349 Special Subsequence 线段树优化DP

　　题目链接：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3349

　　题意：给定一个数列，序列A是一个满足|A_i-A_i-1| <= d的一个序列，其中Ai为给定数列中的元素，要保持相对顺序，求最长的序列A。

　　显然用DP来解决，f[i]表示以第i个数结尾时的最长长度，则f[i]=Max{ f[j]+1 | j<i && |num[j]-num[i]|<=d }。直接转移复杂度O(n^2)，超时。显然对于每个数num只要保存最大的f值就可以了，然后就是查找num[j]在区间[num[i]-d, num[i]+d]的最大的f[j]，用颗线段树维护就可以了。。

 //STATUS:C++_AC_480MS_3008KB
 #include <functional>
 #include <algorithm>
 #include <iostream>
 //#include <ext/rope>
 #include <fstream>
 #include <sstream>
 #include <iomanip>
 #include <numeric>
 #include <cstring>
 #include <cassert>
 #include <cstdio>
 #include <string>
 #include <vector>
 #include <bitset>
 #include <queue>
 #include <stack>
 #include <cmath>
 #include <ctime>
 #include <list>
 #include <set>
 #include <map>
 using namespace std;
 //#pragma comment(linker,"/STACK:102400000,102400000")
 //using namespace __gnu_cxx;
 //define
 #define pii pair<int,int>
 #define mem(a,b) memset(a,b,sizeof(a))
 #define lson l,mid,rt<<1
 #define rson mid+1,r,rt<<1|1
 #define PI acos(-1.0)
 //typedef
 typedef long long LL;
 typedef unsigned long long ULL;
 //const
 const int N=;
 const int INF=0x3f3f3f3f;
 const int MOD=1e+,STA=;
 const LL LNF=1LL<<;
 const double EPS=1e-;
 const double OO=1e15;
 const int dx[]={-,,,};
 const int dy[]={,,,-};
 const int day[]={,,,,,,,,,,,,};
 //Daily Use ...
 inline int sign(double x){return (x>EPS)-(x<-EPS);}
 template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
 template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
 template<class T> inline T lcm(T a,T b,T d){return a/d*b;}
 template<class T> inline T Min(T a,T b){return a<b?a:b;}
 template<class T> inline T Max(T a,T b){return a>b?a:b;}
 template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}
 template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}
 template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}
 template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}
 //End

 int num[N],order[N],hig[N<<],f[N];
 int n,d;

 void update(int l,int r,int rt,int w,int val)
 {
     if(l==r){
         hig[rt]=max(hig[rt],val);
         return ;
     }
     int mid=(l+r)>>;
     if(w<=mid)update(lson,w,val);
     else update(rson,w,val);
     hig[rt]=max(hig[rt<<],hig[rt<<|]);
 }

 int query(int l,int r,int rt,int L,int R)
 {
     if(L<=l && r<=R){
         return hig[rt];
     }
     int mid=(l+r)>>,ret=;
     if(L<=mid)ret=max(ret,query(lson,L,R));
     if(R>mid)ret=max(ret,query(rson,L,R));
     return ret;
 }

 int main()
 {
  //   freopen("in.txt","r",stdin);
     int i,j,m,L,R,lnum,rnum,ans,w;
     while(~scanf("%d%d",&n,&d))
     {
         for(i=;i<n;i++){
             scanf("%d",&num[i]);
             order[i]=num[i];
         }
         sort(order,order+n);
         m=unique(order,order+n)-order;

         mem(hig,);ans=;
         for(i=;i<n;i++){
             lnum=num[i]-d;rnum=num[i]+d;
             L=lower_bound(order,order+m,lnum)-order;
             R=upper_bound(order,order+m,rnum)-order-;
             f[i]=query(,m-,,L,R)+;
             w=lower_bound(order,order+m,num[i])-order;
             update(,m-,,w,f[i]);
             ans=max(ans,f[i]);
         }

         printf("%d\n",ans);
     }
     return ;
 }