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-8;
const double pi = acos(-1.0);
const double inf = ~0u>>2;
const int child_num = 110;
const int BASE = 10000;
const int DIG = 4;
char s[N*100],vir[550];
int id[2024];
struct bignum
{
     int a[110],len;
     bignum()
     {
         memset(a,0,sizeof(a));
         len = 1;
     }
     bignum(int v)
     {
         memset(a,0,sizeof(a));
         len = 0;
         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 = 0 ; i < res.len ;i ++)
         res.a[i] = 0;
         for(i = 0 ; i < res.len ; i++)
         {
             res.a[i] += ((i<len)?a[i]:0)+((i<b.len)?b.a[i]:0);
             res.a[i+1] += res.a[i]/BASE;
             res.a[i] = res.a[i]%BASE;
         }
         if(res.a[res.len]>0) res.len++;
         return res;
     }
     void output()
     {
         printf("%d",a[len-1]);
         for(int i = len-2 ; i >=0 ; i--)
         printf("%04d",a[i]);
         printf("\n");
     }
}dp[110][110];
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[0] = 0;
        //for(int i = 0 ;i < child_num-32 ; i++)
        //id[i+32] = i;
    }
    void reset()
    {
        memset(val,0,sizeof(val));
        memset(fail,0,sizeof(fail));
        memset(ch[0],0,sizeof(ch[0]));
        sz = 1;
    }
    void insert(char *a,int key)
    {
        int k = strlen(a),p = 0;
        for(int i = 0 ; i < k ;i++)
        {
            int d = id[a[i]];
            if(ch[p][d]==0)
            {
                memset(ch[sz],0,sizeof(ch[sz]));
                ch[p][d] = sz++;
            }
            p = ch[p][d];
        }
        val[p] = key;
    }
    void construct(int n)
    {
        int i,head=0,tail = 0;
        for(i = 0; i < n ; i++)
        {
            if(ch[0][i])
            {
                Q[tail++] = ch[0][i];
                fail[ch[0][i]] = 0;
            }
        }
        while(head!=tail)
        {
            int u = Q[head++];
            val[u]|=val[fail[u]];
            for(i = 0 ; 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 = 1; i <= m ;i++)
            for(j = 0 ;j <= sz; j++)
            dp[i][j] = bignum(0);
        dp[0][0] = bignum(1);
        for(i = 0 ; i < m ;i++)
        {
            for(j = 0 ; j < sz ;j++)
            for(g = 0 ; g < n ; g++)
            if(!val[ch[j][g]])
            {
                dp[i+1][ch[j][g]]=dp[i+1][ch[j][g]]+dp[i][j];
            }
        }
        bignum ans = bignum(0);
        for(j = 0 ;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 = 0 ; i < n; i++)
        id[s[i]] = i;
        ac.reset();
        for(i = 1;i <= p; i++)
        {
            scanf("%s",vir);
            ac.insert(vir,1);
        }
        ac.construct(n);
        ac.work(m,n);
    }
    return 0;
}