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

题目链接

难得出一个AC自动机，我还没做到这个题呢。。。这题思路不难想，小小的状压出一维来，不过，D和R，让我wa死了，AC自动机，还得刷啊。。。

 #include<iostream>
 #include<cstring>
 #include<cstdio>
 #include<cmath>
 #include<algorithm>
 using namespace std;
 #define MOD 1000000007
 int dp[][][][];
 int trie[][];
 int o[];
 int fail[],que[],p[][];
 int t;
 void CL()
 {
     t = ;
     memset(trie,-,sizeof(trie));
     memset(o,,sizeof(o));
     memset(p,,sizeof(p));
 }
 void insert(char *s,int x)
 {
     int i,root,len,temp;
     len = strlen(s);
     root = ;
     for(i = ; i < len; i ++)
     {
         temp = s[i] == 'D' ? :;
         if(trie[root][temp] == -)
             trie[root][temp] = t ++;
         root = trie[root][temp];
     }
     o[root] = <<x;
 }
 void build_ac()
 {
     int head,tail,front,i;
     head = tail = ;
     for(i = ; i < ; i ++)
     {
         if(trie[][i] != -)
         {
             fail[trie[][i]] = ;
             que[tail++] = trie[][i];
         }
         else
             trie[][i] = ;
     }
     while(head != tail)
     {
         front = que[head++];
         o[front] |= o[fail[front]];
         for(i = ; i < ; i ++)
         {
             if(trie[front][i] != -)
             {
                 que[tail++] = trie[front][i];
                 fail[trie[front][i]] = trie[fail[front]][i];
             }
             else
             {
                 trie[front][i] = trie[fail[front]][i];
             }
         }
     }
 }
 int main()
 {
     int cas,i,j,k,u,n,m,temp;
     char str[];
     scanf("%d",&cas);
     while(cas--)
     {
         CL();
         scanf("%d%d",&n,&m);
         for(i = ; i < ; i ++)
         {
             scanf("%s",str);
             insert(str,i);
         }
         for(i = ; i <= n; i ++)
         {
             for(j = ; j <= m; j ++)
             {
                 for(k = ; k < t; k ++)
                 {
                     for(u = ; u < ; u ++)
                         dp[i][j][k][u] = ;
                 }
             }
         }
         build_ac();
         dp[][][][] = ;
         for(i = ; i <= n; i ++)
         {
             for(j = ; j <= m; j ++)
             {
                 for(k = ; k < t; k ++)
                 {
                     for(u = ; u < ; u ++)
                     {
                         if(i != n)
                         {
                             temp = trie[k][];
                             dp[i+][j][temp][u|o[temp]] = (dp[i+][j][temp][u|o[temp]] + dp[i][j][k][u])%MOD;
                         }
                         if(j != m)
                         {
                             temp = trie[k][];
                             dp[i][j+][temp][u|o[temp]] = (dp[i][j+][temp][u|o[temp]] + dp[i][j][k][u])%MOD;
                         }
                     }
                 }
             }
         }
         int ans = ;
         for(i = ; i < t; i ++)
         {
             ans = (ans + dp[n][m][i][])%MOD;
         }
         printf("%d\n",ans);
     }
     return ;
 }