Poj 2778 DNA Sequence AC automatic mechanism + matrix rapid Ming

The meaning of the question is very simple. Assume that the text set is A, C, T, G. Given M pattern strings, ask the text whose length is N.
How many possibilities are there...
It is really not intuitive...
The solution is to first learn the AC automatic mechanism and create a Trie graph. Based on the trie graph, we can obtain the path matrix with a length of 1, and then quickly
The path matrix with the length of N is obtained.
It is very hard to understand. I have never learned AC automatic machines. Before learning the AC automatic machine, it is said that you should first learn the Trie tree and KMP
To understand. It took nearly two days to learn about the AC automatic machine Trie diagram, and it was another day to understand this question.
The competition is coming soon. I don't know whether it is good or bad to switch from Changchun to Jinhua... It's still a weak dish...
Paste my Trie graph + quick nether (directly binary, not the algorithm written in number theory )...

# Include <stdio. h>
# Include <string. h>
# Include <algorithm>
# Include <queue>
Using namespace std;

Typedef long INT;
Const int MOD = 100000;
Const int MAX_P = 100;
Const int MAX_D = 4;
Int nIdx [256];
Char szPat [MAX_P];
INT nMatrix [MAX_P] [MAX_P];
Int B [MAX_P] [MAX_P];
Int a [MAX_P] [MAX_P];

Void InitIdx ()
{
NIdx ['a'] = 0;
NIdx ['C'] = 1;
NIdx ['T'] = 2;
NIdx ['G'] = 3;
}

Struct Trie
{
Trie * fail;
Trie * next [MAX_D];
Int no;
Bool flag;
Trie ()
{
Fail = NULL;
Memset (next, 0, sizeof (next ));
No = 0;
Flag = false;
}
};
Trie tries [MAX_D * MAX_P];
Int nP;
Trie * pRoot;

Trie * NewNode ()
{
Memset (& tries [nP], 0, sizeof (Trie ));
Tries [nP]. no = nP;
Return & tries [nP ++];
}

Void InitTrie (Trie * & pRoot)
{
NP = 0;
PRoot = NewNode ();
}

Void Insert (char * pszPat)
{
Trie * pNode = pRoot;

While (* pszPat)
{
If (pNode-> next [nIdx [* pszPat] = NULL)
{
PNode-> next [nIdx [* pszPat] = NewNode ();
}
PNode = pNode-> next [nIdx [* pszPat];
++ PszPat;
}
PNode-> flag = true;
}

Int BuildAC (Trie * pRoot)
{
Memset (nMatrix, 0, sizeof (nMatrix ));

PRoot-> fail = NULL;
Queue <Trie *> qt;
Qt. push (pRoot );
While (! Qt. empty ())
{
Trie * front = qt. front ();
Qt. pop ();

For (int I = 0; I <MAX_D; ++ I)
{
If (front-> next [I])
{
Trie * pNode = front-> fail;
While (pNode & pNode-> next [I] = NULL)
{
PNode = pNode-> fail;
}
Front-> next [I]-> fail = pNode? PNode-> next [I]: pRoot;
If (front-> next [I]-> fail-> flag = true)
{
Front-> next [I]-> flag = true;
}

Qt. push (front-> next [I]);
}
Else
{
Front-> next [I] = front = pRoot? PRoot: front-> fail-> next [I];
}

If (front-> next [I]-> flag = false)
{
NMatrix [front-> no] [front-> next [I]-> no] ++;
}
}
}

Return nP; // total number of nodes
}

Void MultyMatrix (int a [] [MAX_P], int B [] [MAX_P], INT C [] [MAX_P], int nSize)
{
For (int I = 0; I <nSize; ++ I)
{
For (int j = 0; j <nSize; ++ j)
{
INT nSum = 0;
For (int k = 0; k <nSize; ++ k)
{
NSum = (nSum + A [I] [k] * B [k] [j]) % MOD;
}
C [I] [j] = nSum;
}
}
}

Void CopyMatrix (int a [] [MAX_P], int B [] [MAX_P], int nSize)
{
For (int I = 0; I <nSize; ++ I)
{
For (int j = 0; j <nSize; ++ j)
{
A [I] [j] = B [I] [j];
}
}
}

Void MatrixPower (int m [] [MAX_P], int nSize, INT nP)
{
If (nP = 1)
{
CopyMatrix (A, M, nSize );
Return;
}

MatrixPower (M, nSize, nP/2 );
MultyMatrix (A, A, B, nSize );
If (nP % 2)
{
MultyMatrix (B, M, A, nSize );
}
Else
{
CopyMatrix (A, B, nSize );
}
}

Int main ()
{
INT nM, nN;

InitIdx ();
While (scanf ("% I64d % I64d", & nM, & nN) = 2)
{
InitTrie (pRoot );
While (nM --)
{
Scanf ("% s", szPat );
Insert (szPat );
}
Int nSize = BuildAC (pRoot );

MatrixPower (nMatrix, nSize, nN );
INT nAns = 0;
For (int I = 0; I <nSize; ++ I)
{
NAns = (nAns + A [0] [I]) % MOD;
}
Printf ("% I64d \ n", nAns % MOD );
}

Return 0;
}