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=100010;
const int INF=0x3f3f3f3f;
const int MOD=1e+7,STA=8000010;
const LL LNF=1LL<<60;
const double EPS=1e-8;
const double OO=1e15;
const int dx[4]={-1,0,1,0};
const int dy[4]={0,1,0,-1};
const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
//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<<2],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)>>1;
    if(w<=mid)update(lson,w,val);
    else update(rson,w,val);
    hig[rt]=max(hig[rt<<1],hig[rt<<1|1]);
}

int query(int l,int r,int rt,int L,int R)
{
    if(L<=l && r<=R){
        return hig[rt];
    }
    int mid=(l+r)>>1,ret=0;
    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=0;i<n;i++){
            scanf("%d",&num[i]);
            order[i]=num[i];
        }
        sort(order,order+n);
        m=unique(order,order+n)-order;

        mem(hig,0);ans=0;
        for(i=0;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-1;
            f[i]=query(0,m-1,1,L,R)+1;
            w=lower_bound(order,order+m,num[i])-order;
            update(0,m-1,1,w,f[i]);
            ans=max(ans,f[i]);
        }

        printf("%d
",ans);
    }
    return 0;
}