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！

  1 //STATUS:C++_AC_281MS_3768KB
  2 #include <functional>
  3 #include <algorithm>
  4 #include <iostream>
  5 //#include <ext/rope>
  6 #include <fstream>
  7 #include <sstream>
  8 #include <iomanip>
  9 #include <numeric>
 10 #include <cstring>
 11 #include <cassert>
 12 #include <cstdio>
 13 #include <string>
 14 #include <vector>
 15 #include <bitset>
 16 #include <queue>
 17 #include <stack>
 18 #include <cmath>
 19 #include <ctime>
 20 #include <list>
 21 #include <set>
 22 #include <map>
 23 using namespace std;
 24 //using namespace __gnu_cxx;
 25 //define
 26 #define pii pair<int,int>
 27 #define mem(a,b) memset(a,b,sizeof(a))
 28 #define lson l,mid,rt<<1
 29 #define rson mid+1,r,rt<<1|1
 30 #define PI acos(-1.0)
 31 //typedef
 32 typedef __int64 LL;
 33 typedef unsigned __int64 ULL;
 34 //const
 35 const int N=100010;
 36 //const LL INF=0x3f3f3f3f;
 37 //const int MOD=1000000007,STA=8000010;
 38 const LL LNF=1LL<<60;
 39 const double EPS=1e-8;
 40 const double OO=1e15;
 41 const int dx[4]={-1,0,1,0};
 42 const int dy[4]={0,1,0,-1};
 43 const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
 44 //Daily Use ...
 45 inline int sign(double x){return (x>EPS)-(x<-EPS);}
 46 template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
 47 template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
 48 template<class T> inline T lcm(T a,T b,T d){return a/d*b;}
 49 template<class T> inline T Min(T a,T b){return a<b?a:b;}
 50 template<class T> inline T Max(T a,T b){return a>b?a:b;}
 51 template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}
 52 template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}
 53 template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}
 54 template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}
 55 //End
 56 
 57 int num[N];
 58 int fup[N],fdown[N],dup[N],ddown[N];
 59 int wup[N],wdown[N],cntup[N],cntdown[N];
 60 int upk,downk;
 61 int T,n;
 62 
 63 int search_up(int *d,int l,int r,int tar)
 64 {
 65     int mid;
 66     while(l<r){
 67         mid=(l+r)>>1;
 68         if(d[mid]<=tar)l=mid+1;
 69         else r=mid;
 70     }
 71     return l;
 72 }
 73 
 74 int search_down(int *d,int l,int r,int tar)
 75 {
 76     int mid;
 77     while(l<r){
 78         mid=(l+r)>>1;
 79         if(d[mid]>=tar)l=mid+1;
 80         else r=mid;
 81     }
 82     return l;
 83 }
 84 
 85 int main()
 86 {
 87  //   freopen("in.txt","r",stdin);
 88     int i,j,hig,w;
 89     scanf("%d",&T);
 90     while(T--)
 91     {
 92         scanf("%d",&n);
 93         for(i=1;i<=n;i++){
 94             scanf("%d",&num[i]);
 95         }
 96         upk=downk=2;
 97         dup[1]=0x7fffffff,ddown[1]=0x80000000;
 98         hig=1;
 99         for(i=n;i>=1;i--){
100             w=search_up(dup,1,upk,num[i]);
101             if(w==1 || num[i]!=dup[w-1])cntup[i]=1;
102             else cntup[i]=cntup[wup[w-1]]+1;
103             dup[w]=num[i];wup[w]=i;
104             fup[i]=w;
105             upk=Max(upk,w+1);
106             w=search_down(ddown,1,downk,num[i]);
107             if(w==1 || num[i]!=ddown[w-1])cntdown[i]=1;
108             else cntdown[i]=cntdown[wdown[w-1]]+1;
109             ddown[w]=num[i];wdown[w]=i;
110             fdown[i]=w;
111             downk=Max(downk,w+1);
112 
113             hig=Max(hig,fup[i]+fdown[i]-Min(cntup[i],cntdown[i]));
114         }
115 
116         printf("%d
",hig);
117     }
118     return 0;
119 }