Poj1_1 -- DNA repair (AC automation + dp), poj1_1 -- dna Automation

Poj1_1 -- DNA repair (AC automation + dp), poj1_1 -- dna Automation
DNA repair

Time Limit:2000 MS		Memory Limit:65536 K
Total Submissions:5743		Accepted:2693

Description

Biologists finally invent techniques of repairing DNA that contains segments causing kinds of inherited diseases. for the sake of simplicity, a DNA is represented as a string containing characters 'A', 'G', 'C' and 'T '. the repairing techniques are simply to change some characters to eliminate all segments causing diseases. for example, we can repair a DNA "AAGCAG" to "AGGCAC" to eliminate the initial causing disease segments "AAG", "AGC" and "CAG" by changing two characters. note that the retried red DNA can still contain only characters 'A', 'G', 'C' and 'T '.

You are to help the biologists to repair a DNA by changing least number of characters.

Input

The input consists of multiple test cases. Each test case starts with a line containing one integers N(1 ≤ N≤ 50), which is the number of DNA segments causing inherited diseases.
The following NLines gives NNon-empty strings of length not greater than 20 containing only characters in "AGCT", which are the DNA segments causing inherited disease.
The last line of the test case is a non-empty string of length not greater than 1000 containing only characters in "AGCT", which is the DNA to be retried red.

The last test case is followed by a line containing one zeros.

Output

For each test case, print a line containing the test case number (beginning with 1) followed by
Number of characters which need to be changed. If it's impossible to repair the given DNA, print-1.

Sample Input

2AAAAAGAAAG    2ATGTGAATG4AGCTAGT0

Sample Output

Case 1: 1Case 2: 4Case 3: -1

Perform automatic machine construction on the diseased string given. dp [I] [j] corresponds to the minimum value that takes step I to reach the j node from the root.

 

#include <cstdio>#include <cstring>#include <algorithm>#include <queue>using namespace std ;#define INF 0x3f3f3f3fstruct node{    int flag , id ;    node *next[4] , *fail ;} tree[2100] ;queue <node *> que ;int num , dp[1100][2100] ;char str[2100] ;char c[5] = "ACGT" ;node *newnode(){    node *p = &tree[num] ;    p->flag = 0 ;    p->id = num++ ;    p->fail = NULL ;    for(int i = 0 ; i < 4 ; i++)        p->next[i] = NULL ;    return p ;}void settree(char *s,node *rt){    int i , k , l = strlen(s) ;    node *p = rt ;    for(i = 0 ; i < l ; i++)    {        for(k = 0 ; k < 4 ; k++)            if( s[i] == c[k] )                break ;        if( p->next[k] == NULL )            p->next[k] = newnode() ;        p = p->next[k] ;    }    p->flag = 1 ;    return ;}void setfail(node *rt){    node *p = rt , *temp ;    p->fail = NULL;    while( !que.empty() ) que.pop() ;    que.push(p) ;    while( !que.empty() )    {        p = que.front() ;        que.pop() ;        for(int i = 0 ; i < 4 ; i++)        {            if( p->next[i] )            {                temp = p->fail ;                while( temp && !temp->next[i] )                    temp = temp->fail ;                p->next[i]->fail = temp ? temp->next[i] : rt ;                que.push(p->next[i]) ;                if( temp && temp->flag )                    p->flag = 1 ;            }            else                p->next[i] = p == rt ? rt : p->fail->next[i] ;        }    }    return ;}int query(char *s,node *rt){    int i , j , k , l = strlen(s) , flag ;    memset(dp,INF,sizeof(dp)) ;    dp[0][0] = 0 ;    for(i = 0 ; i < l ; i++)    {        for(j = 0 ; j < num ; j++)        {            for(k = 0 ; k < 4 ; k++)            {                if( tree[j].next[k]->flag ) continue ;                if( s[i] == c[k] )                    dp[i+1][ tree[j].next[k]->id ] = min( dp[i][j] , dp[i+1][ tree[j].next[k]->id ] ) ;                else                    dp[i+1][ tree[j].next[k]->id ] = min( dp[i][j]+1 , dp[i+1][ tree[j].next[k]->id ] ) ;            }        }        /*for(j = 0 ; j < num ; j++)            printf("%d ", dp[i+1][j]) ;        printf("\n") ;*/    }    int ans = INF ;    for(i = 0 ; i < num ; i++)        ans = min(ans,dp[l][i]) ;    if( ans == INF )        ans = -1 ;    return ans ;}int main(){    int i , n , temp = 1 ;    node *rt ;    while( scanf("%d", &n) && n )    {        num = 0 ;        rt = newnode() ;        for(i = 0 ; i < n ; i++)        {            scanf("%s", str) ;            settree(str,rt) ;        }        setfail(rt) ;        scanf("%s", str) ;        printf("Case %d: %d\n", temp++ , query(str,rt) ) ;    }    return 0 ;}