Poj2533 longest ordered subsequence

Question:
Pure longest ascending subsequence (LIS: longest increasing subsequence)
State transition equation (time complexity is O (n * n ))
If (A [I] <A [J])
DP [I] = max (DP [I], DP [J] + 1 );

Solution 1:
Based on the state transition equation:
If (A [I] <A [J])
DP [I] = max (DP [I], DP [J] + 1 );

# Include <iostream>
# Include <cstring>
Using namespace STD;
# Define x 1003
Int DP [x];
Int A [x];
Int max (int A, int B)
{
Return A> B? A: B;
}
Int main ()
{
Freopen ("sum. In", "r", stdin );
Freopen ("sum. Out", "W", stdout );
Int N;

While (CIN> N)
{
Memset (A, 0, sizeof ());
For (INT I = 1; I <= N; I ++)
Cin> A [I];
For (INT I = N; I> = 1; I --)
For (Int J = I + 1; j <= N; j ++)
If (A [I] <A [J])
DP [I] = max (DP [I], DP [J] + 1); // CoreAlgorithm
Int max = 0;
For (INT I = 0; I <= N; I ++) // find the largest ascending subsequence
If (max <DP [I])
Max = DP [I];
Cout <MAX + 1 <Endl;
}
Return 0;
}

Solution 2:
O (N ^ 2) algorithm analysis is as follows: (Num [1]... num [N] stores all input numbers)
1. For num [N], because it is the last number, when you start to search for num [N], there is only a child sequence with a length of 1 and no descent;
2. If you start searching for num [n-1], there are two possibilities:
(1) If num [n-1] <num [N], there will be a child sequence num [n-1], num [N] with a length of 2.
(2) If num [n-1]> num [N], there will be a child sequence num [n-1] Or num [N] with a length of 1.
3. Generally, if the subsequence starts from num [J], the maximum length of the subsequence is not decreased at this time, which should be obtained using the following method:
In a [J + 1], a [J + 2],... in a [n], find a child sequence that is larger than a [T] and is the longest and does not drop as its successor.
4. For algorithm requirements, two arrays are defined: num [Max] Memory number Lis [Max] Maximum ascending sub-sequence length starting from I

You can also:
D: array [1. N, 1. 3] of integer;
D [T, 1] indicates a [T]
D [T, 2] indicates the maximum length of the subsequence from the I position to n.
D [T, 3] indicates the next position of the subsequence starting from position I.

# Include <iostream>
# Include <cstring>
Using namespace STD;
# Define x 1003
Int DP [x];
Int A [x];
Int max (int A, int B)
{
Return A> B? A: B;
}
Int main ()
{
Freopen ("sum. In", "r", stdin );
Freopen ("sum. Out", "W", stdout );
Int N;

While (CIN> N)
{
Memset (A, 0, sizeof ());
For (INT I = 1; I <= N; I ++)
Cin> A [I];
Int ans = 0;
For (INT I = N; I> = 1; I --)
{
Int max = 0;
For (Int J = I + 1; j <= N; j ++)
If (A [I] <A [J] & DP [J]> MAX)
Max = DP [J];
DP [I] = MAX + 1;
If (DP [I]> ans)
Ans = DP [I];
}
Cout <ans <Endl;
}

Return 0;
}