UVA 11468(Substring-AC自动机上dp)[Template:AC自动机]

Substring

Given
a set of pattern strings, and a text, you have to find, if any of the pattern is a substring of the text. If any of the pattern string can be found in text, then print “yes”, otherwise “no” (without quotes).

But,
unfortunately, that’s not what is asked here. J

The
problem described above, requires a input file generator. The generator generates a text of length L, by choosing L characters randomly. Probability of choosing each character is given as priori, and independent of choosing others.

Now,
given a set of patterns, calculate the probability of a valid program generating “no”.

Input

First
line contains an integer T, the number of test cases. Each case starts with an integer K, the number of pattern strings. Next K lines each contain a pattern string, followed by an integer N, number of valid characters. Next N lines
each contain a character and the probability of selecting that character, p_i. Next an integer L, the length of the string generated. The generated text can consist of only the valid characters, given above.

There
will be a blank line after each test case.

Output

For
each test case, output the number of test case, and the probability of getting a “no”.

Constraints

· T ≤
50

· K
≤ 20

· Length
of each pattern string is between 1 and 20

· Each
pattern string consists of only alphanumeric characters (‘a’ to ‘z’, ‘A’ to ‘Z’,’0’ to ‘9’)

· Valid
characters are all alphanumeric characters

· ∑p_i =
1

· L
≤ 100

Sample
input

a
0.5

b
0.5

a
0.2

b
0.8

Output
for Sample Input

Case
#1: 0.250000

Case
#2: 0.840000

本题利用了AC自动机

首先，把模式串p推入AC自动机，然后dp[i][j]表示匹配到i点没成功，还剩j个字符未匹配的概率，dp

注意AC自动机上的dp（我之前忘了getFail()了，，，考挂自己弱）

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<cctype>
#include<ctime>
#include<queue>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p])
#define Lson (x<<1)
#define Rson ((x<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (100000007)
#define MAXT (50+10)
#define MAXL (100+10)
#define MAXN (20+10)
#define MAXLen (20+10)
#define MAXNode (400+10)
#define Sigma_size (62)
typedef long long ll;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return (a-b+(a-b)/F*F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}

int n,L;
char s[MAXLen];
double prob[Sigma_size],dp[MAXNode][MAXL];
bool b[MAXNode][MAXL]; 

class Aho_Corasick_Automata
{
public:
	int ch[MAXNode][Sigma_size];
	int v[MAXNode],siz;
	// AC自动机
	int f[MAXNode],last[MAXNode];
	Aho_Corasick_Automata(int _siz=0):siz(_siz){MEM(ch) MEM(v) MEM(f) MEM(last)}
	void mem(int _siz=0){siz=_siz; MEM(ch) MEM(v) MEM(f) MEM(last)	}
	int idx(char c){return 'a'<=c&&c<='z'?c-'a':'A'<=c&&c<='Z'?c-'A'+26:c-'0'+52;}
	void insert(char *s,int val=0)
	{
		int u=0,n=strlen(s);
		Rep(i,n)
		{
			int c=idx(s[i]);
			if (!ch[u][c])
			{
				++siz;
				MEM(ch[siz]);
				ch[u][c]=siz;
			}
			u=ch[u][c];
		}
		v[u]=val;
	}
	void getFail()
	{
		queue<int> q;
		Rep(c,Sigma_size)
		{
			int u=ch[0][c];
			if (u) q.push(u),last[u]=0;
		}
		while (!q.empty())
		{
			int r=q.front();q.pop();  //r--c-->u
			Rep(c,Sigma_size)
			{
				int u=ch[r][c];
				if (!u) {ch[r][c]=ch[f[r]][c]; continue;}
				q.push(u);
				f[u]=ch[f[r]][c];
				last[u]=v[f[u]]?f[u]:last[f[u]];
			}
		}
	}
	void print(int j) //打印全串中所有以j为末尾的str
	{
		if (j)
		{
			printf("%d %d\n",j,v[j]);
			print(last[j]);
		}
	}
	void find(char *s)
	{
		int u=0,n=strlen(s);
		Rep(i,n)
		{
			int c=idx(s[i]);
			u=ch[u][c];
			if (v[u]) print(u);
			else if (last[u]) print(u);
		}
	}

	double dfs(int u,int l)
	{
		if (!l) return 1;
		if (b[u][l]) return dp[u][l];
		double ans=0.0;
		Rep(c,Sigma_size)
		if (prob[c])
		{
			if (v[ch[u][c]]||last[ch[u][c]]) continue;
			ans+=prob[c]*dfs(ch[u][c],l-1);
		}	

		b[u][l]=1;
		return dp[u][l]=ans;
	}

}T;

int main()
{
//	freopen("uva11468.in","r",stdin);
//	freopen(".out","w",stdout);
	int kcase;
	scanf("%d",&kcase);
	For(tt,kcase)
	{
		T.mem();
		MEM(dp) MEM(prob) MEM(b)

		scanf("%d",&n);
		For(i,n)
		{
			scanf("%s",s);
			T.insert(s,1);
		}
		T.getFail(); 

		scanf("%d\n",&n);
		For(i,n)
		{
			char c;
			scanf("%c",&c);
			scanf("%lf\n",&prob[T.idx(c)]);
		}

		scanf("%d",&L);

		printf("Case #%d: %.6lf\n",tt,T.dfs(0,L));
	}

	return 0;
}

时间： 2024-10-07 09:18:28

UVA 11468(Substring-AC自动机上dp)[Template:AC自动机]

Input

Output

Output
for Sample Input