HDU 4758 Walk Through Squares( AC自动机 + 状态压缩DP )

题意：给你两个串A,B, 问一个串长为M+N且包含A和B且恰好包含M个R的字符串有多少种组合方式，所有字符串中均只含有字符L和R。

dp[i][j][k][S]表示串长为i，有j个R，在自动机中的状态为k，包含AB的状态为S的方案个数。

PS1.之前用long long int超时了两次

PS2.把行列搞错了WA了几次

#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cstring>

//#define LL long long int

using namespace std;

const int MAX_NODE = 1210;
const int CHILD_NUM = 2;
const int MAXN = 122;
const int MOD = 1000000007;

struct ACAutomaton
{
int chd[MAX_NODE][CHILD_NUM]; //每个节点的儿子,即当前节点的状态转移
int val[MAX_NODE];            //记录题目给的关键数据
int fail[MAX_NODE];           //传说中的fail指针
int Q[MAX_NODE<<1];           //队列,用于广度优先计算fail指针
int ID[128];                  //字母对应的ID
int sz;                       //已使用节点个数

//初始化,计算字母对应的儿子ID,如:'a'->0 ... 'z'->25
void Initialize()
{
fail[0] = 0;
ID['L'] = 0;
ID['R'] = 1;
return;
}
//重新建树需先Reset
void Reset()
{
memset(chd[0] , 0 , sizeof(chd[0]));
val[0] = 0;
sz = 1;
}
//将权值为key的字符串a插入到trie中
void Insert(char *a,int key)
{
int p = 0;
for ( ; *a ; a ++)
{
int c = ID[*a];
if (!chd[p][c])
{
memset(chd[sz] , 0 , sizeof(chd[sz]));
val[sz] = 0;
chd[p][c] = sz ++;
}
p = chd[p][c];
}
val[p] = key;
}
//建立AC自动机,确定每个节点的权值以及状态转移
void Construct()
{
int *s = Q , *e = Q;
for (int i = 0 ; i < CHILD_NUM ; i ++)
{
if (chd[0][i])
{
fail[ chd[0][i] ] = 0;
*e ++ = chd[0][i];
}
}
while (s != e)
{
int u = *s++;
for (int i = 0 ; i < CHILD_NUM ; i ++)
{
int &v = chd[u][i];
if (v)
{
*e ++ = v;
fail[v] = chd[ fail[u] ][i];
//以下一行代码要根据题目所给val的含义来写
val[v] |= val[ fail[v] ];
}
else
{
v = chd[ fail[u] ][i];
}
}
}
}
} AC;

int M, N;
char str[MAXN];
int dp[2][MAXN][MAXN<<1][1<<2];

void solved()
{
int pre = 0, cur = 1;

memset( dp[0], 0, sizeof(dp[0]) );
dp[0][0][0][0] = 1;
int all = 1 << 2;
int L = M + N;

for ( int i = 0; i < L; ++i )
{
memset( dp[cur], 0, sizeof(dp[cur]) );
for ( int j = 0; j <= M; ++j )
for ( int k = 0; k < AC.sz; ++k )
for ( int r = 0; r < 2; ++r )
for ( int S = 0; S < all; ++S )
{
int next = AC.chd[k][r];
dp[cur][j+r][next][ S|AC.val[next] ] += dp[pre][j][k][S];
dp[cur][j+r][next][ S|AC.val[next] ] %= MOD;
}

cur ^= 1;
pre ^= 1;
}

int ans = 0;
for ( int j = 0; j < AC.sz; ++j )
{
ans += dp[pre][M][j][all-1];
ans %= MOD;
}
printf( "%d\n", ans );

return;
}

int main()
{
AC.Initialize();
int T;
scanf( "%d", &T );
while ( T-- )
{
scanf( "%d%d", &M, &N );
AC.Reset();
for ( int i = 0; i < 2; ++i )
{
scanf( "%s", str );
AC.Insert( str, 1 << i );
}
AC.Construct();

solved();
}
return 0;
}