南阳214----单调递增子序列(二)

 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
",s);
49     }
50     return 0;
51 }