BZOJ 1264 [AHOI2006]基因匹配Match

 1 //F[i]表示B与A中前i个数的LCS长度
 2  
 3 #include <cstdio>
 4 #include <algorithm>
 5 
 6 using namespace std;
 7 
 8 void Get_Val(int &Ret)
 9 {
10     Ret = 0;
11     char ch;
12     while (ch = getchar(), ch > ‘9‘ || ch < ‘0‘)
13         ;
14     do
15     {
16         (Ret *= 10) += ch - ‘0‘;
17     }
18     while (ch = getchar(), ch >= ‘0‘ && ch <= ‘9‘);
19 }
20 
21 const int Max_N(20050);
22 
23 int N;
24 int A[Max_N * 5], B[Max_N * 5];
25 int tot[Max_N], g[Max_N][6];
26 
27 int F[Max_N * 5];
28 int BIT[Max_N * 5];
29 
30 inline
31 int lowbit(const int &x)
32 {
33     return x & -x;
34 }
35 
36 void Change(int i, const int &x)
37 {
38     while (i <= N * 5)
39     {
40         BIT[i] = max(BIT[i], x);
41         i += lowbit(i);
42     }
43 }
44 
45 int Query(int i)
46 {
47     int Ret(0);
48     while (i)
49     {
50         Ret = max(Ret, BIT[i]);
51         i -= lowbit(i);
52     }
53     return Ret;
54 }
55 
56 void init()
57 {
58     Get_Val(N);
59     for (int i = 1;i <= N * 5;++i)
60         Get_Val(A[i]), g[A[i]][++tot[A[i]]] = i;
61     for (int i = 1;i <= N * 5;++i)
62         Get_Val(B[i]);
63 }
64 
65 void dp()
66 {
67     int k, Ans(0);
68     for (int i = 1;i <= N * 5;++i)
69         for (int p = 5;p >= 1;--p)
70         {
71             k = g[B[i]][p];
72             Ans = max(Ans, F[k] = Query(k - 1) + 1);
73             Change(k, F[k]);
74         }
75     printf("%d\n", Ans);
76 }
77 
78 int main()
79 {
80     init();
81     dp();
82     return 0;
83 }