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
", 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;
}