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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”.
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.
There will be a blank line after each test case.
For each test case, output the number of test case, and the probability of getting a “no”.
Constraints
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Sample input
2
1
a
2
a 0.5
b 0.5
2
2
ab
ab
2
a 0.2
b 0.8
2
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; }
UVA 11468(Substring-AC自动机上dp)[Template:AC自动机]
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原文地址:http://blog.csdn.net/nike0good/article/details/44133199