ZOJ-3349 Special Subsequence 线段树优化DP

时间：2023-03-09 05:30:38
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|Ai-Ai-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 ;

 }

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