poj1625 Censored! AC自动机+DP-优快云博客

本文链接：https://blog.youkuaiyun.com/corncsd/article/details/39211241

本文深入探讨了Censored!问题的解决方案，包括输入参数解释、算法设计与实现细节，重点在于通过构造AC自动机来优化查找过程，最终计算出在特定约束下可使用的不同句子数量。使用动态规划方法解决该问题，涉及大数运算和状态转移方程的应用。

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Censored!

Time Limit: 5000MS		Memory Limit: 10000K
Total Submissions: 8108		Accepted: 2193

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 1
ab
bb

Sample Output

这跟上面那个题是同一个意思，找长度为M并且不包含给的串，这样的串的个数。

但是这个答案会很大，要用大数，所以矩阵快速幂不好弄，用DP做。

先构造好AC自动机后，一开始在节点0，用dp[i][j]表示走i步到达节点j的路径数，dp[i][j]=Σdp[i-1][k]，k能一步走到j，并且是合法的。这样最后答案就是Σdp[M][i]，0<=i<ac.sz。

也可以像上面那个题那样先构造一步矩阵，dp[i][j]=dp[i-1][k]*mat[k][j]，k到j是合法的。

#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include<vector>
#include<cmath>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<algorithm>
#define MAXNODE 110
using namespace std;
int N,M,K,dp[55][110][110],ans[110],SIGMA_SIZE;
char str[55];
map<char,int> mp;
struct AC{
    int ch[MAXNODE][60],f[MAXNODE],val[MAXNODE],sz;
    void init(){
        memset(ch[0],0,sizeof(ch[0]));
        val[0]=0;
        sz=1;
    }
    int idx(char c){
        return mp[c];
    }
    void insert(char *s,int v){
        int u=0;
        for(int i=0;s[i];i++){
            int c=idx(s[i]);
            if(!ch[u][c]){
                memset(ch[sz],0,sizeof(ch[sz]));
                val[sz]=0;
                ch[u][c]=sz++;
            }
            u=ch[u][c];
        }
        val[u]=1;
    }
    void get_fail(){
        queue<int> q;
        f[0]=0;
        for(int c=0;c<SIGMA_SIZE;c++){
            int u=ch[0][c];
            if(u){
                f[u]=0;
                q.push(u);
            }
        }
        while(!q.empty()){
            int r=q.front();
            q.pop();
            for(int c=0;c<SIGMA_SIZE;c++){
                int u=ch[r][c];
                if(!u){
                    ch[r][c]=ch[f[r]][c];
                    continue;
                }
                q.push(u);
                f[u]=ch[f[r]][c];
                val[u]|=val[f[u]];
            }
        }
    }
}ac;
void add(int *a,int *b){
    int c=0;
    for(int i=0;i<100;i++){
        a[i]+=b[i]+c;
        c=a[i]/10;
        a[i]%=10;
    }
}
void DP(){
    memset(dp,0,sizeof(dp));
    dp[0][0][0]=1;
    for(int i=1;i<=M;i++)
        for(int j=0;j<ac.sz;j++) if(!ac.val[j]){
                for(int k=0;k<N;k++) if(!ac.val[ac.ch[j][k]]) add(dp[i][ac.ch[j][k]],dp[i-1][j]);
        }
    memset(ans,0,sizeof(ans));
    for(int i=0;i<ac.sz;i++) add(ans,dp[M][i]);
    int p;
    for(p=99;p>=0&&!ans[p];p--);
    if(p<0) printf("0\n");
    else{
        for(;p>=0;p--) printf("%d",ans[p]);
        puts("");
    }
}
int main(){
    freopen("in.txt","r",stdin);
    while(scanf("%d%d%d",&N,&M,&K)!=EOF){
        scanf("%s",str);
        mp.clear();
        int k=0;
        for(int i=0;str[i];i++) mp[str[i]]=k++;
        SIGMA_SIZE=k;
        ac.init();
        while(K--){
            scanf("%s",str);
            ac.insert(str,1);
        }
        ac.get_fail();
        DP();
    }
    return 0;
}