1 /*
2 用一个数组记录递增子序列，保持s处为最长子序列的最后一个值
3 当输入x小于d[s]时，向前找x的位置覆盖即可
4 复杂度与经典算法同为n*n
5 加入二分查找，优化后为n*logn
6 */
7 #include<cstdio>
8 #define inf 1<<30
9 int d[100005],s;
10
11 void solven(int x)
12 {
13 int t,left,right,mid;
14 if(s == 0 || d[s] < x)
15 d[++s] = x;
16 else
17 {
18 left = 1;
19 right = s;
20 while(left <= right)
21 {
22 mid = (left + right)/2;
23 if(d[mid]>=x && d[mid-1]<x)
24 {
25 d[mid] = x;
26 return ;
27 }
28 if(d[mid] > x)
29 right = mid - 1;
30 else
31 left = mid + 1;
32 }
33 }
34 }
35
36 int main()
37 {
38 int n,x,i,t;
39 while(~scanf("%d",&n))
40 {
41 s = 0;
42 d[0] = -inf;
43 while(n--)
44 {
45 scanf("%d",&x);
46 solven(x);
47 }
48 printf("%d\n",s);
49 }
50 return 0;
51 }
