• 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 }
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  • 原文地址:https://www.cnblogs.com/zhsl/p/3209558.html
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