HDU-4604 Deque DP

　　题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4604

　　因为deque最后的数列是单调不降的，因此，我们可以枚举数列中的某个中间数Ai，如果从中间数Ai开始，如果后面的要和这个中间数形成单调不降的序列，那么后面的数必须是单调不降或者单调不升的序列，才能进入deque中，因此为两者长度的和，这就是一个LIS的DP。然后枚举的时候从后往前枚举，复杂度O( n*log n)。这里要注意一点，存在相同元素，因此需要减去两个里面出现Ai次数的最小值！

 //STATUS:C++_AC_281MS_3768KB
 #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;
 //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 __int64 LL;
 typedef unsigned __int64 ULL;
 //const
 const int N=;
 //const LL INF=0x3f3f3f3f;
 //const int MOD=1000000007,STA=8000010;
 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];
 int fup[N],fdown[N],dup[N],ddown[N];
 int wup[N],wdown[N],cntup[N],cntdown[N];
 int upk,downk;
 int T,n;

 int search_up(int *d,int l,int r,int tar)
 {
     int mid;
     while(l<r){
         mid=(l+r)>>;
         if(d[mid]<=tar)l=mid+;
         else r=mid;
     }
     return l;
 }

 int search_down(int *d,int l,int r,int tar)
 {
     int mid;
     while(l<r){
         mid=(l+r)>>;
         if(d[mid]>=tar)l=mid+;
         else r=mid;
     }
     return l;
 }

 int main()
 {
  //   freopen("in.txt","r",stdin);
     int i,j,hig,w;
     scanf("%d",&T);
     while(T--)
     {
         scanf("%d",&n);
         for(i=;i<=n;i++){
             scanf("%d",&num[i]);
         }
         upk=downk=;
         dup[]=0x7fffffff,ddown[]=0x80000000;
         hig=;
         for(i=n;i>=;i--){
             w=search_up(dup,,upk,num[i]);
             if(w== || num[i]!=dup[w-])cntup[i]=;
             else cntup[i]=cntup[wup[w-]]+;
             dup[w]=num[i];wup[w]=i;
             fup[i]=w;
             upk=Max(upk,w+);
             w=search_down(ddown,,downk,num[i]);
             if(w== || num[i]!=ddown[w-])cntdown[i]=;
             else cntdown[i]=cntdown[wdown[w-]]+;
             ddown[w]=num[i];wdown[w]=i;
             fdown[i]=w;
             downk=Max(downk,w+);

             hig=Max(hig,fup[i]+fdown[i]-Min(cntup[i],cntdown[i]));
         }

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