hdu4758Walk Through Squares(ac自动机+dp）

链接

dp[x][y][node][sta] 表示走到在x,y位置node节点时状态为sta的方法数，因为只有2个病毒串，这时候的状态只有4种，根据可走的方向转移一下。

这题输入的是m、N,先列后行，因为输反了，WA了N次啊。。

 #include <iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<stdlib.h>
 #include<vector>
 #include<cmath>
 #include<queue>
 #include<set>
 using namespace std;
 #define N 210
 #define LL long long
 #define INF 0xfffffff
 const double eps = 1e-;
 const double pi = acos(-1.0);
 const double inf = ~0u>>;
 const int child_num = ;
 const int mod = ;
 int n,m;
 int dis[][] = {,,,};
 char vir[];
 class AC
 {
     private:
     int ch[N][child_num];
     int fail[N];
     int val[N];
     int Q[N];
     int id[];
     int dp[N][][][];
     int sz;
     public:
     void init()
     {
         fail[] = ;
         id['R'] = ;
         id['D'] = ;
     }
     void reset()
     {
         memset(val,,sizeof(val));
         memset(ch[],,sizeof(ch[]));
         sz = ;
     }
     void insert(char *a,int key)
     {
         int p = ;
         for( ; *a ; a++)
         {
             int d = id[*a];
             if(ch[p][d]==)
             {
                 memset(ch[sz],,sizeof(ch[sz]));
                 ch[p][d] = sz++;
             }
             p = ch[p][d];
         }
         val[p] = (<<(key-));
     }
     void construct()
     {
         int i,head = ,tail = ;
         for(i =  ;  i < child_num;  i++)
         {
             if(ch[][i])
             {
                 fail[ch[][i]] = ;
                 Q[tail++] = ch[][i];
             }
         }
         while(tail!=head)
         {
             int u = Q[head++];
             val[u]|=val[fail[u]];
             for(i = ; i < child_num ; i++)
             {
                 if(ch[u][i])
                 {
                     fail[ch[u][i]] = ch[fail[u]][i];
                     Q[tail++] = ch[u][i];
                 }
                 else ch[u][i] = ch[fail[u]][i];
             }
         }
     }
     void work()
     {
         int i,j,g;
         for(i = ;i <= sz ;i++)
             for(j = ; j <= n+ ;j++)
                 for(g = ; g <= m+; g++)
                 for(int e = ; e  <  ; e++)
                 dp[i][j][g][e] = ;
         //memset(dp,0,sizeof(dp));

         dp[][][][] = ;
         for(j = ; j <= n ;j ++)
             for(g = ; g <= m ; g++)
             {
                 for(i =  ;i < sz ; i++)
                 for(int o = ; o <  ; o++)
                 for(int e =  ;e < child_num ;e++)
                 {
                     int tx = dis[e][]+j;
                     int ty = dis[e][]+g;
                     int p = ch[i][e];
                     int pp = val[ch[i][e]];
                     dp[p][tx][ty][o|pp] = (dp[p][tx][ty][o|pp]+dp[i][j][g][o])%mod;
                 }
             }
         int ans = ;
         for(i =  ;i < sz ;i++)
         ans = (ans+dp[i][n][m][])%mod;
         printf("%d\n",ans%mod);
     }
 }ac;
 int main()
 {
     int t;
     cin>>t;
     ac.init();
     while(t--)
     {
         ac.reset();
         scanf("%d%d",&n,&m);
         n++,m++;
         scanf("%s",vir);
         ac.insert(vir,);
         scanf("%s",vir);
         ac.insert(vir,);
         ac.construct();
         ac.work();
     }
     return ;
 }