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=100010;
//const LL INF=0x3f3f3f3f;
//const int MOD=1000000007,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];
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)>>1;
        if(d[mid]<=tar)l=mid+1;
        else r=mid;
    }
    return l;
}

int search_down(int *d,int l,int r,int tar)
{
    int mid;
    while(l<r){
        mid=(l+r)>>1;
        if(d[mid]>=tar)l=mid+1;
        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=1;i<=n;i++){
            scanf("%d",&num[i]);
        }
        upk=downk=2;
        dup[1]=0x7fffffff,ddown[1]=0x80000000;
        hig=1;
        for(i=n;i>=1;i--){
            w=search_up(dup,1,upk,num[i]);
            if(w==1 || num[i]!=dup[w-1])cntup[i]=1;
            else cntup[i]=cntup[wup[w-1]]+1;
            dup[w]=num[i];wup[w]=i;
            fup[i]=w;
            upk=Max(upk,w+1);
            w=search_down(ddown,1,downk,num[i]);
            if(w==1 || num[i]!=ddown[w-1])cntdown[i]=1;
            else cntdown[i]=cntdown[wdown[w-1]]+1;
            ddown[w]=num[i];wdown[w]=i;
            fdown[i]=w;
            downk=Max(downk,w+1);

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

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