[Longest non-ascending subsequence + nlog (n)] Peking University POJ 3636 Nested Dolls

[Cpp]
/* The program is made by pyy */
/*----------------------------------------------------------------------------//
Copyright (c) 2012 panyanyany All rights reserved.
 
URL: http://poj.org/problem? Id = 3636
Name: 3636 Nested Dolls-longest non-ascending subsequence
 
Date: Monday, July 9, 2012
Time Stage: two hours
 
Result:
10404790 panyanyany
3636
Accepted 400 K 157 ms c ++
1827B 11:01:39
 
Test Data:
 
Review
Www.2cto.com
//----------------------------------------------------------------------------*/
 
# Include <stdio. h>
# Include <stdlib. h>
# Include <string. h>
# Include <math. h>
# Include <vector>
 
# Include <algorithm>
# Include <iostream>
# Include <queue>
# Include <set>
# Include <string>
 
Using namespace std;
 
# Define MEM (a, v) memset (a, v, sizeof (a) // a for address, v for value
# Define max (x, y) (x)> (y )? (X): (y ))
# Define min (x, y) (x) <(y )? (X): (y ))
 
# Define INF (0x3f3f3f)
# Define maxn20009
 
# Define L (x) <1)
# Define R (x) <1) | 1)
# Define M (x, y) (x) + (y)> 1)
 
# Define DB //
 
Struct NODE {
Int w, h;
};
 
Bool cmp (const NODE & lhs, const NODE & rhs)
{
If (lhs. w = rhs. w)
Return lhs. h> rhs. h;
Return lhs. w <rhs. w;
}
 
Int order [MAXN];
NODE a [MAXN];
 
Int LNotIncS (NODE a [], int n)
{
Int I, r, l, len, m;
 
MEM (order, 0 );
Sort (a, a + n, cmp );
 
Len = 1;
 
For (I = 0; I <n; ++ I)
{
R = len;
L = 1;
While (l <= r)
{
M = (l + r)> 1;
If (order [m]> = a [I]. h)
{
L = m + 1;
}
Else
R = m-1;
}
 
// The length of the updated ordered table is the maximum value of L.
// If the ordered table is ascending, the minimum value is updated.
If (order [l] <a [I]. h)
Order [l] = a [I]. h;
Len = max (len, l );
}
 
Return len;
}
 
Int main ()
{
Int I, tcase, n;
While (scanf ("% d", & tcase )! = EOF)
{
While (tcase --)
{
Scanf ("% d", & n );
For (I = 0; I <n; ++ I)
{
Scanf ("% d", & a [I]. w, & a [I]. h );
}
Printf ("% d \ n", LNotIncS (a, n ));
}
}
Return 0;
}
Author: panyanyany