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。。
1 #include <iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 #include<stdlib.h> 6 #include<vector> 7 #include<cmath> 8 #include<queue> 9 #include<set> 10 using namespace std; 11 #define N 210 12 #define LL long long 13 #define INF 0xfffffff 14 const double eps = 1e-8; 15 const double pi = acos(-1.0); 16 const double inf = ~0u>>2; 17 const int child_num = 2; 18 const int mod = 1000000007; 19 int n,m; 20 int dis[2][2] = {1,0,0,1}; 21 char vir[150]; 22 class AC 23 { 24 private: 25 int ch[N][child_num]; 26 int fail[N]; 27 int val[N]; 28 int Q[N]; 29 int id[128]; 30 int dp[N][110][110][4]; 31 int sz; 32 public: 33 void init() 34 { 35 fail[0] = 0; 36 id['R'] = 0; 37 id['D'] = 1; 38 } 39 void reset() 40 { 41 memset(val,0,sizeof(val)); 42 memset(ch[0],0,sizeof(ch[0])); 43 sz = 1; 44 } 45 void insert(char *a,int key) 46 { 47 int p =0 ; 48 for( ; *a ; a++) 49 { 50 int d = id[*a]; 51 if(ch[p][d]==0) 52 { 53 memset(ch[sz],0,sizeof(ch[sz])); 54 ch[p][d] = sz++; 55 } 56 p = ch[p][d]; 57 } 58 val[p] = (1<<(key-1)); 59 } 60 void construct() 61 { 62 int i,head = 0,tail = 0; 63 for(i = 0 ; i < child_num; i++) 64 { 65 if(ch[0][i]) 66 { 67 fail[ch[0][i]] = 0; 68 Q[tail++] = ch[0][i]; 69 } 70 } 71 while(tail!=head) 72 { 73 int u = Q[head++]; 74 val[u]|=val[fail[u]]; 75 for(i = 0; i < child_num ; i++) 76 { 77 if(ch[u][i]) 78 { 79 fail[ch[u][i]] = ch[fail[u]][i]; 80 Q[tail++] = ch[u][i]; 81 } 82 else ch[u][i] = ch[fail[u]][i]; 83 } 84 } 85 } 86 void work() 87 { 88 int i,j,g; 89 for(i = 0;i <= sz ;i++) 90 for(j = 1; j <= n+1 ;j++) 91 for(g = 1; g <= m+1; g++) 92 for(int e =0 ; e < 4 ; e++) 93 dp[i][j][g][e] = 0; 94 //memset(dp,0,sizeof(dp)); 95 96 dp[0][1][1][0] = 1; 97 for(j = 1; j <= n ;j ++) 98 for(g = 1; g <= m ; g++) 99 { 100 for(i = 0 ;i < sz ; i++) 101 for(int o = 0; o < 4 ; o++) 102 for(int e = 0 ;e < child_num ;e++) 103 { 104 int tx = dis[e][0]+j; 105 int ty = dis[e][1]+g; 106 int p = ch[i][e]; 107 int pp = val[ch[i][e]]; 108 dp[p][tx][ty][o|pp] = (dp[p][tx][ty][o|pp]+dp[i][j][g][o])%mod; 109 } 110 } 111 int ans = 0; 112 for(i = 0 ;i < sz ;i++) 113 ans = (ans+dp[i][n][m][3])%mod; 114 printf("%d ",ans%mod); 115 } 116 }ac; 117 int main() 118 { 119 int t; 120 cin>>t; 121 ac.init(); 122 while(t--) 123 { 124 ac.reset(); 125 scanf("%d%d",&n,&m); 126 n++,m++; 127 scanf("%s",vir); 128 ac.insert(vir,1); 129 scanf("%s",vir); 130 ac.insert(vir,2); 131 ac.construct(); 132 ac.work(); 133 } 134 return 0; 135 }