Poj 1036 Gangsters

POJ's 2nd questions. It is also a dynamic conversion. dp [I] [j] indicates the maximum value of the I-th moment and the degree to which the door is opened to j.

Of course, the status is quite good. The value of this moment must be related to the previous moment. The connection is the j of the previous moment, that is, the degree of the door of the previous moment, and there are only three possibilities for a time gate. 1: open a unit. 2: close a unit. 3: Same as before. In addition, we need to use a rolling array to do this. Otherwise, my memory may be too large. I have never tried this kind of dynamic transfer equation. Generally, we use a rolling array to save space. It's just a matter of course. The following code is provided.

[Cpp]
# Include <iostream>
# Include <algorithm>
Using namespace std;
Struct node {
Int time;
Int value;
Int silm;
};
Node gangster [110];
Int mark [30010];
Int dp [3] [2, 110];
Bool cmp (node a, node B)
{
If (a. time <B. time)
Return true;
Else return false;
}
Int maxi (int a, int B)
{
If (a> B)
Return;
Else return B;
}
Int main ()
{
Int N, K, T, I, j, w, p, mabi;
Cin> N> K> T;
For (I = 1; I <= N; I ++)
{
Cin> gangster [I]. time;
Mark [gangster [I]. time] = 1;
}
For (I = 1; I <= N; I ++)
Cin> gangster [I]. value;
For (I = 1; I <= N; I ++)
Cin> gangster [I]. silm;
Sort (gangster + 1, gangster + 1 + N, cmp );
Mabi = 0;
For (I = 0; I <= T; I ++)
For (j = 0; j <= I & j <= K; j ++)
{
W = 0;
Dp [I % 2] [j] = 0;
If (mark [I] = 1)
{
For (p = 1; p <= N; p ++)
{
If (gangster [p]. time = I & gangster [p]. silm = j)
W = w + gangster [p]. value;
}
}
If (j = 0)
{
Dp [I % 2] [j] = maxi (dp [1-i % 2] [j], dp [1-i % 2] [j + 1]);
}
Else if (j = K)
{
Dp [I % 2] [j] = maxi (dp [1-i % 2] [j], dp [1-i % 2] [J-1]);
}
Else dp [I % 2] [j] = maxi (dp [1-i % 2] [j], dp [1-i % 2] [J-1]), dp [1-i % 2] [j + 1]);
Dp [I % 2] [j] + = w;
If (dp [I % 2] [j]> mabi)
Mabi = dp [I % 2] [j];
}
Cout <mabi <endl;
Return 0;
}

# Include <iostream>
# Include <algorithm>
Using namespace std;
Struct node {
Int time;
Int value;
Int silm;
};
Node gangster [110];
Int mark [30010];
Int dp [3] [2, 110];
Bool cmp (node a, node B)
{
If (a. time <B. time)
Return true;
Else return false;
}
Int maxi (int a, int B)
{
If (a> B)
Return;
Else return B;
}
Int main ()
{
Int N, K, T, I, j, w, p, mabi;
Cin> N> K> T;
For (I = 1; I <= N; I ++)
{
Cin> gangster [I]. time;
Mark [gangster [I]. time] = 1;
}
For (I = 1; I <= N; I ++)
Cin> gangster [I]. value;
For (I = 1; I <= N; I ++)
Cin> gangster [I]. silm;
Sort (gangster + 1, gangster + 1 + N, cmp );
Mabi = 0;
For (I = 0; I <= T; I ++)
For (j = 0; j <= I & j <= K; j ++)
{
W = 0;
Dp [I % 2] [j] = 0;
If (mark [I] = 1)
{
For (p = 1; p <= N; p ++)
{
If (gangster [p]. time = I & gangster [p]. silm = j)
W = w + gangster [p]. value;
}
}
If (j = 0)
{
Dp [I % 2] [j] = maxi (dp [1-i % 2] [j], dp [1-i % 2] [j + 1]);
}
Else if (j = K)
{
Dp [I % 2] [j] = maxi (dp [1-i % 2] [j], dp [1-i % 2] [J-1]);
}
Else dp [I % 2] [j] = maxi (dp [1-i % 2] [j], dp [1-i % 2] [J-1]), dp [1-i % 2] [j + 1]);
Dp [I % 2] [j] + = w;
If (dp [I % 2] [j]> mabi)
Mabi = dp [I % 2] [j];
}
Cout <mabi <endl;
Return 0;
}
This program should be able to optimize the time complexity, which was previously done. It may be supplemented later.