POJ 1625 ac自动机+高精度dp

Censored!

Time Limit: 5000MS Memory Limit: 10000KTotal Submissions: 7469 Accepted: 2023

Description

The alphabet of Freeland consists of exactly N letters. Each sentence of Freeland language (also known as Freish) consists of exactly M letters without word breaks. So, there exist exactly N^M different Freish sentences.

But after recent election of Mr. Grass Jr. as Freeland president some words offending him were declared unprintable and all sentences containing at least one of them were forbidden. The sentence S contains a word W if W is a substring of S i.e. exists such k >= 1 that S[k] = W[1], S[k+1] = W[2], ...,S[k+len(W)-1] = W[len(W)], where k+len(W)-1 <= M and len(W) denotes length of W. Everyone who uses a forbidden sentence is to be put to jail for 10 years.

Find out how many different sentences can be used now by freelanders without risk to be put to jail for using it.

Input

The first line of the input file contains three integer numbers: N -- the number of letters in Freish alphabet, M -- the length of all Freish sentences and P -- the number of forbidden words (1 <= N <= 50, 1 <= M <= 50, 0 <= P <= 10).

The second line contains exactly N different characters -- the letters of the Freish alphabet (all with ASCII code greater than 32).

The following P lines contain forbidden words, each not longer than min(M, 10) characters, all containing only letters of Freish alphabet.

Output

Output the only integer number -- the number of different sentences freelanders can safely use.

Sample Input

2 3 1abbb

Sample Output

5

题意：给定n个字符，p个禁止出现的序列，求长度为m的合法序列有多少个。

解题思路很简单，不过这个题计算结果不是取模，只能高精度，wa多次，tle，mle都出现了，过得比较艰难。

代码：

/* ***********************************************Author :xianxingwuguanCreated Time :2014-2-5 11:29:37File Name :3.cpp************************************************ */#pragma comment(linker, "/STACK:102400000,102400000")#include <stdio.h>#include <iostream>#include <algorithm>#include <sstream>#include <stdlib.h>#include <string.h>#include <limits.h>#include <string>#include <time.h>#include <math.h>#include <queue>#include <stack>#include <set>#include <map>using namespace std;#define INF 0x3f3f3f3f#define eps 1e-8#define pi acos(-1.0)typedef long long ll;map<char,int> mp;int n;struct BigInt{    const static int mod = 10000;    const static int DLEN = 4;    int a[110],len;    BigInt()    {        memset(a,0,sizeof(a));        len = 1;    }    BigInt(int v)    {        memset(a,0,sizeof(a));        len = 0;        do        {            a[len++] = v%mod;            v /= mod;        }while(v);    }    BigInt(const char s[])    {        memset(a,0,sizeof(a));        int L = strlen(s);        len = L/DLEN;        if(L%DLEN)len++;        int index = 0;        for(int i = L-1;i >= 0;i -= DLEN)        {            int t = 0;            int k = i - DLEN + 1;            if(k < 0)k = 0;            for(int j = k;j <= i;j++)                t = t*10 + s[j] - '0';            a[index++] = t;        }    }    BigInt operator +(const BigInt &b)const    {        BigInt res;        res.len = max(len,b.len);        for(int i = 0;i <= res.len;i++)            res.a[i] = 0;        for(int 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]/mod;            res.a[i] %= mod;        }        if(res.a[res.len] > 0)res.len++;        return res;    }    BigInt operator *(const BigInt &b)const    {        BigInt res;        for(int i = 0; i < len;i++)        {            int up = 0;            for(int j = 0;j < b.len;j++)            {                int temp = a[i]*b.a[j] + res.a[i+j] + up;                res.a[i+j] = temp%mod;                up = temp/mod;            }            if(up != 0)                res.a[i + b.len] = up;        }        res.len = len + b.len;        while(res.a[res.len - 1] == 0 &&res.len > 1)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];struct Trie{int next[110][70],end[110],fail[110];int root,L;int newnode(){for(int i=0;i<n;i++)next[L][i]=-1;end[L++]=0;return L-1;}void init(){L=0;root=newnode();}void insert(char *str){int len=strlen(str);int now=root;for(int i=0;i<len;i++){int p=mp[str[i]];if(next[now][p]==-1)next[now][p]=newnode();now=next[now][p];}end[now]|=1;}void build(){queue<int> q;fail[root]=root;for(int i=0;i<n;i++)if(next[root][i]==-1)next[root][i]=root;else {fail[next[root][i]]=root;q.push(next[root][i]);}while(!q.empty()){int now=q.front();q.pop();end[now]|=end[fail[now]];for(int i=0;i<n;i++)if(next[now][i]==-1)next[now][i]=next[fail[now]][i];else {fail[next[now][i]]=next[fail[now]][i];q.push(next[now][i]);}}}void solve(int m){for(int i=0;i<=m;i++)for(int j=0;j<=L;j++)dp[i][j]=BigInt(0);dp[0][0]=BigInt(1);for(int i=0;i<m;i++)for(int j=0;j<L;j++){for(int k=0;k<n;k++){int p=next[j][k];if(end[p])continue;dp[i+1][p]=dp[i][j]+dp[i+1][p];}}BigInt ans=BigInt(0);    for(int i=0;i<L;i++)ans=ans+dp[m][i];ans.output();}}ac;char str[10000];int main(){     //freopen("data.in","r",stdin);     //freopen("data.out","w",stdout);     int i,j,k,m,p; while(~scanf("%d%d%d",&n,&m,&p)){ ac.init();mp.clear(); getchar(); gets(str); for(i=0;i<n;i++) mp[str[i]]=i; while(p--){ gets(str); ac.insert(str); } ac.build(); ac.solve(m); }     return 0;}