Hoj 3008 Matryoshka Dolls Again largest independent subset

 

/*

 

 

 

Question:

 

When the length, width, and height of a doll are larger than the other, the small one can be put in and asked about

 

Minimum Number

 

 

 

Analysis:

 

Find the largest independent set, use the total number of dolls-the maximum match, and use two layers of loops to create a picture first. Every time the length, width, and height of a doll are larger than j hours,

 

Then g [I] [j] =
If it is true, you can create a graph. At the beginning, I first sorted it, and then I found that it was not needed... Direct

 

You can create a graph by repeating the two layers and then comparing them. Finding the maximum match is a pure hungy algorithm.

 

 

 

*/

 

# Include
<Iostream>

 

# Include
<Cstring>

 

# Include
<Cstdio>

 

Using namespace
Std;

 

# Define X 502

 

Int
Xm [X], ym [X], n, m;

 

Bool g [X] [X], use [X];

 

Struct node

 

{

 

Int a, B, c;

 

} P [X];

 

Bool dfs (int u)

 

{

 

For (int v = 1; v <= n; v ++)

 

If (g [u] [v] &! Use [v])

 

{

 

Use [v] = true;

 

If (ym [v] =-1 | dfs (ym [v])

 

{

 

Xm [u] = v;

 

Ym [v] = u;

 

Return true;

 

}

 

}

 

Return false;

 

}

 

Int hungry ()

 

{

 

Memset (xm,-1, sizeof (xm ));

 

Memset (ym,-1, sizeof (ym ));

 

Int ret = 0;

 

For (int u = 1; u <= n; u ++)

 

If (xm [u] =-1)

 

{

 

Memset (use, false, sizeof (use ));

 

If (dfs (u ))

 

Ret ++;

 

}

 

Return ret;

 

}

 

Int main ()

 

{

 

Freopen ("sum. in", "r", stdin );

 

Freopen ("sum. out", "w", stdout );

 

While (cin> n, n)

 

{

 

For (int I = 1; I <= n; I ++)

 

Scanf ("% d", & p [I]. a, & p [I]. B, & p [I]. c );

 

Memset (g, false, sizeof (g ));

 

For (int I = 1; I <= n; I ++) // create a graph

 

For (int j = 1; j <= n; j ++)

 

If (I! = J & p [j]. a> p [I]. a & p [j]. b> p [I]. B & p [j]. c> p [I]. c) // when all three conditions are met

 

G [j] [I] = true;

 

Printf ("% d \ n", n-hungry ());

 

}

 

 

 

Return 0;

 

}