poj1625Censored!(AC自动机+dp）

链接

第一次做这种题目，参考了下题解，相当于把树扯直了做DP，估计这一类题都是这个套路吧。

状态方程dp[i][next] = dp[i][next]+dp[i][j] ;dp[i][j]表示长度为i的第J个结点的时候满足题意的num，next为当前j点所能走到的下一个合法的结点。

需要用高精度，看到一些规范的高精度写法，觉得不错，有空整理下来。

不知道是不是我理解错了，按理说字符串病毒长度不应超过10.。但开到55依旧RE，开550AC。。。

 #include <iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<stdlib.h>
 #include<vector>
 #include<cmath>
 #include<queue>
 #include<set>
 using namespace std;
 #define N 110
 #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 BASE = ;
 const int DIG = ;
 char s[N*],vir[];
 int id[];
 struct bignum
 {
      int a[],len;
      bignum()
      {
          memset(a,,sizeof(a));
          len = ;
      }
      bignum(int v)
      {
          memset(a,,sizeof(a));
          len = ;
          do
          {
              a[len++] = v%BASE;
              v/=BASE;
          }while(v);
      }
      /*bignum(const char s[])
      {
          memset(a,0,sizeof(a));
          int k = strlen(s);
          len = k/DIG;
          if(k%DIG) len++;
          int cnt = 0;
          for(int i = k-1;  i >= 0 ; i-=DIG)
          {
              int t = 0;
              int kk = i-DIG+1;
              if(kk<0) kk =0;
              for(int j = kk ; j <= i ; j++)
              t = t*10+s[j]-'0';
              a[cnt++] = t;
          }
      }*/
      bignum operator + (const bignum &b)const
      {
          bignum res;
          res.len = max(len,b.len);
          int i;
          for(i =  ; i < res.len ;i ++)
          res.a[i] = ;
          for(i =  ; i < res.len ; i++)
          {
              res.a[i] += ((i<len)?a[i]:)+((i<b.len)?b.a[i]:);
              res.a[i+] += res.a[i]/BASE;
              res.a[i] = res.a[i]%BASE;
          }
          if(res.a[res.len]>) res.len++;
          return res;
      }
      void output()
      {
          printf("%d",a[len-]);
          for(int i = len- ; i >= ; i--)
          printf("%04d",a[i]);
          printf("\n");
      }
 }dp[][];
 class AC
 {
     private:
     int ch[N][child_num];
     int Q[N];
     int val[N];
     int fail[N];
     //int id[N];
     int sz;
     public :
     void init()
     {
         fail[] = ;
         //for(int i = 0 ;i < child_num-32 ; i++)
         //id[i+32] = i;
     }
     void reset()
     {
         memset(val,,sizeof(val));
         memset(fail,,sizeof(fail));
         memset(ch[],,sizeof(ch[]));
         sz = ;
     }
     void insert(char *a,int key)
     {
         int k = strlen(a),p = ;
         for(int i =  ; i < k ;i++)
         {
             int d = id[a[i]];
             if(ch[p][d]==)
             {
                 memset(ch[sz],,sizeof(ch[sz]));
                 ch[p][d] = sz++;
             }
             p = ch[p][d];
         }
         val[p] = key;
     }
     void construct(int n)
     {
         int i,head=,tail = ;
         for(i = ; i < n ; i++)
         {
             if(ch[][i])
             {
                 Q[tail++] = ch[][i];
                 fail[ch[][i]] = ;
             }
         }
         while(head!=tail)
         {
             int u = Q[head++];
             val[u]|=val[fail[u]];
             for(i =  ; i < n ; i++)
             {
                 if(ch[u][i])
                 {
                     Q[tail++] = ch[u][i];
                     fail[ch[u][i]] = ch[fail[u]][i];
                 }
                 else ch[u][i] = ch[fail[u]][i];
             }
         }
     }
     void work(int m,int n)
     {
         int i,j,g;
         for(i = ; i <= m ;i++)
             for(j =  ;j <= sz; j++)
             dp[i][j] = bignum();
         dp[][] = bignum();
         for(i =  ; i < m ;i++)
         {
             for(j =  ; j < sz ;j++)
             for(g =  ; g < n ; g++)
             if(!val[ch[j][g]])
             {
                 dp[i+][ch[j][g]]=dp[i+][ch[j][g]]+dp[i][j];
             }
         }
         bignum ans = bignum();
         for(j =  ;j < sz ; j++)
         ans=ans+dp[m][j];
         ans.output();
     }
 }ac;
 int main()
 {
     int n,m,i,p;
     ac.init();
     while(cin>>n>>m>>p)
     {
         cin>>s;
         for(i =  ; i < n; i++)
         id[s[i]] = i;
         ac.reset();
         for(i = ;i <= p; i++)
         {
             scanf("%s",vir);
             ac.insert(vir,);
         }
         ac.construct(n);
         ac.work(m,n);
     }
     return ;
 }