poj 1625 Censored!//AC自动机+DP+大数

最新推荐文章于 2022-02-25 19:38:17 发布

HQD因为有趣所以做题

最新推荐文章于 2022-02-25 19:38:17 发布

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分类专栏： ACM_数据结构文章标签： tree input integer c numbers each

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本文介绍了一种基于矩阵乘法的算法，用于计算在给定的字母表和禁止词汇条件下，能够安全使用的句子数量。通过构建有限状态机并利用矩阵快速幂的方法，有效地解决了这一问题。

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Censored!

Time Limit: 5000MS		Memory Limit: 10000K
Total Submissions: 3608		Accepted: 999

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

Source

Northeastern Europe 2001, Northern Subregion

/*
用矩阵乘法没跑过去的代码
用数组RE,用MAP就TLE。。。。。。
我2了
*/
#include<cstdio>
#include<cstring>
#include<map>
using namespace std;
const int NODE=110;
const int MAXN=25;
const int mod=10000;
int CH;
int tree[NODE][50],cnt;
int Que[NODE];
int word[NODE];
int fail[NODE];
char str[101];
int matnum,matcal[NODE];
map<char ,int>num;
struct bint
{
    int dig[MAXN],len;
    bint()
    {
        dig[0]=0;
        len=0;
    }
} mm[NODE][NODE],r[NODE][NODE],te[NODE][NODE];
void print(bint& a)
{
    int i;
    i=a.len-1;
    printf("%d",a.dig[a.len]);
    while(i>=0)  printf("%04d",a.dig[i--]);
    printf("/n");
}
bint add(bint a,int b)
{
    int i;
    bint c;
    for(i=0; i<=a.len||b; ++i)
    {
        if(i<=a.len)  b+=a.dig[i];
        c.dig[i]=b%mod;
        b/=mod;
    }
    c.len=i-1;
    return c;
}
bint add(bint a,bint b)
{
    int i,carry;
    bint c;
    for(i=carry=0; i<=a.len||i<=b.len||carry; ++i)
    {
        if(i<=a.len)  carry+=a.dig[i];
        if(i<=b.len)  carry+=b.dig[i];
        c.dig[i]=carry%mod;
        carry/=mod;
    }
    c.len=i-1;
    return c;
}
bint by(bint a,bint b)
{
    int i,j,carry;
    bint c;
    for(i=a.len+b.len+1; i>=0; i--)  c.dig[i]=0;
    for(i=0; i<=a.len; i++)
    {
        carry=0;
        for(j=0; j<=b.len||carry; ++j)
        {
            carry+=c.dig[i+j];
            if(j<=b.len)  carry+=a.dig[i]*b.dig[j];
            c.dig[i+j]=carry%mod;
            carry/=mod;
        }
    }
    i=a.len+b.len+1;
    while(i&&c.dig[i]==0)  i--;
    c.len=i;
    return c;
}
void Ins(char *a)
{
    int p=0;
    for(; *a; a++)
    {
        int c=num[*a];
        if(!tree[p][c])
        {
            memset(tree[cnt],0,sizeof(tree[cnt]));
            word[cnt]=0;
            tree[p][c]=cnt++;
        }
        p=tree[p][c];
    }
    word[p]=1;
}
void Input(int n,int p)
{
    fail[0]=0;
    memset(tree[0],0,sizeof(tree[0]));
    cnt=1;
    scanf("%s",str);
    for(int i=0; i<n; ++i)  num[str[i]]=i;
    for(int i=1; i<=p; ++i)
    {
        scanf("%s",str);
        Ins(str);
    }
}
void AC()
{
    int *s=Que,*e=Que;
    for(int i=0; i<CH; ++i)
        if(tree[0][i])
        {
            fail[tree[0][i]]=0;
            *e++=tree[0][i];
        }
    while(s!=e)
    {
        int p=*s++;
        for(int i=0; i<CH; ++i)
        {
            if(tree[p][i])
            {
                int v=tree[p][i];
                *e++=v;
                fail[v]=tree[fail[p]][i];
                word[v]|=word[fail[v]];
            }
            else
            {
                tree[p][i]=tree[fail[p]][i];
            }
        }
    }
}
void work()
{
    matnum=0;
    memset(matcal,0,sizeof(matcal));
    memset(mm,0,sizeof(mm));
    for(int i=0; i<cnt; ++i)
        if(word[i]==0)  matcal[i]=matnum++;
    for(int i=0; i<cnt; ++i)
        if(word[i]==0)
        {
            for(int j=0; j<CH; ++j)
            {
                if(word[tree[i][j]]==0)
                    mm[matcal[i]][matcal[tree[i][j]]]=add(mm[matcal[i]][matcal[tree[i][j]]],1);
            }
        }
}
void Mul(bint a[][NODE],bint b[][NODE])
{
    memset(te,0,sizeof(te));
    for(int i=0; i<matnum; ++i)
        for(int j=0; j<matnum; ++j)
        {
            bint tt;
            for(int k=0; k<matnum; ++k)
                tt=add(tt,by(a[i][k],b[k][j]));
            te[i][j]=tt;
        }
}
void give(int a,bint& b)
{
    b.dig[0]=a%mod;
    a/=mod;
    if(a>0)  b.dig[1]=a,b.len=1;
    else b.len=0;
}
void solve(int n)
{
    for(int i=0; i<matnum; ++i)  give(1,r[i][i]);
    while(n)
    {
        if(n&1)
        {
            Mul(mm,r);
            memcpy(r,te,sizeof(te));
        }
        n=n>>1;
        if(n)
        {
            Mul(mm,mm);
            memcpy(mm,te,sizeof(te));
        }
    }
    bint res;
    for(int i=0; i<matnum; i++)  //res+=r[0][i];
        res=add(res,r[0][i]);
    print(res);
    //printf("%I64d/n",res);
}
int main()
{
    int n,m,p;
    scanf("%d%d%d",&n,&m,&p);
    CH=n;
    Input(n,p);
    AC();
    work();
    solve(m);
    return 0;
}

/* 用数组RE。。。。 */ #include<cstdio> #include<cstring> #include<map> using namespace std; #define swap(a,b) {int t=a;a=b;b=t;} const int NODE=110; const int MAXN=50; const int mod=1000000000; int CH; int tree[NODE][51],cnt; int Que[NODE]; int word[NODE]; int fail[NODE]; char str[101]; //int num[300]; map<char ,int >num; int n,m,p; struct bint { int dig[MAXN],len; bint() { dig[0]=0; len=0; } }; bint count[2][NODE]; int x,y;//作滚动数组用 bool vis[NODE]; void print(bint& a) { int i; i=a.len-1; printf("%d",a.dig[a.len]); while(i>=0) printf("%09d",a.dig[i--]); printf("/n"); } bint add(bint a,bint b) { int i,carry; bint c; for(i=carry=0; i<=a.len||i<=b.len||carry; ++i) { if(i<=a.len) carry+=a.dig[i]; if(i<=b.len) carry+=b.dig[i]; c.dig[i]=carry%mod; carry/=mod; } c.len=i-1; return c; } void give(int a,bint& b) { b.dig[0]=a%mod; a/=mod; if(a>0) b.dig[1]=a,b.len=1; else b.len=0; } void Ins(char *a) { int p=0; for(; *a; a++) { int c=num[*a]; if(!tree[p][c]) { memset(tree[cnt],0,sizeof(tree[cnt])); word[cnt]=0; tree[p][c]=cnt++; } p=tree[p][c]; } word[p]=1; } void Input() { fail[0]=0; memset(tree[0],0,sizeof(tree[0])); cnt=1; scanf("%s",str); for(int i=0; i<n; ++i) num[str[i]]=i; for(int i=1; i<=p; ++i) { scanf("%s",str); Ins(str); } } void AC() { int *s=Que,*e=Que; for(int i=0; i<CH; ++i) if(tree[0][i]) { fail[tree[0][i]]=0; *e++=tree[0][i]; } while(s!=e) { int p=*s++; for(int i=0; i<CH; ++i) { if(tree[p][i]) { int v=tree[p][i]; *e++=v; fail[v]=tree[fail[p]][i]; word[v]|=word[fail[v]]; } else { tree[p][i]=tree[fail[p]][i]; } } } } bint ans; void dfs(int idx) { if(vis[idx]) return ; vis[idx]=1; for(int i=0;i<CH;++i) if(!word[tree[idx][i]]) { count[x][tree[idx][i]]=add(count[x][tree[idx][i]],count[y][idx]); dfs(tree[idx][i]); } } void dfs_ans(int idx) { if(vis[idx]) return ; vis[idx]=1; ans=add(ans,count[x][idx]); for(int i=0;i<CH;++i) if(!word[tree[idx][i]]) dfs_ans(tree[idx][i]); } void solve() { give(1,count[1][0]); x=1,y=0; for(int i=1;i<=m;i++) { swap(x,y);//做滚动数组用 memset(vis,0,sizeof(vis));//用了虚拟儿子 memset(count[x],0,sizeof(count[y])); dfs(0);//暴力N次 } memset(vis,0,sizeof(vis)); dfs_ans(0); print(ans); } int main() { scanf("%d%d%d",&n,&m,&p); CH=n; Input(); AC(); solve(); return 0; }