最长上升子序列nlogn解法详解 poj 2533

先上一个n^2的算法：

 1 #include <iostream>
 2 using namespace std;
 3
 4 const int N = 1000;
 5 int a[N];
 6 int g[N];
 7
 8 int main ()
 9 {
10     int n;
11     while ( cin >> n )
12     {
13         for ( int i = 0; i < n; i++ )
14         {
15             cin >> a[i];
16             g[i] = 1;
17         }
18         int ans = g[0];
19         for ( int i = 1; i < n; i++ )
20         {
21             for ( int j = 0; j < i; j++ )
22             {
23                 if ( a[j] >= a[i] ) continue;
24                 if ( g[j] + 1 > g[i] ) g[i] = g[j] + 1;
25             }
26             if ( g[i] > ans ) ans = g[i];
27         }
28         cout << ans << endl;
29     }
30     return 0;
31 }

然后是nlogn的：

 1 #include <algorithm>
 2 #include <iostream>
 3 using namespace std;
 4
 5 const int INF = 1 << 29;
 6 const int N = 1000;
 7 int s[N];
 8 int top;
 9
10 int main ()
11 {
12     int n;
13     while ( cin >> n )
14     {
15         top = 0;
16         s[top++] = INF;
17         for ( int i = 0; i < n; i++ )
18         {
19             int tmp;
20             cin >> tmp;
21             if ( tmp > s[top - 1] )
22             {
23                 s[top++] = tmp;
24             }
25             else
26             {
27                 int pos = upper_bound( s, s + top, tmp ) - s;
28                 s[pos] = tmp;
29             }
30         }
31         cout << top << endl;
32     }
33     return 0;
34 }