DNA repair
| Time Limit: 2000MS |
|
Memory Limit: 65536K |
| 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 repaired 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
N lines gives
N non-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 repaired.
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 the
number of characters which need to be changed. If it's impossible to repair the given DNA, print -1.
Sample Input
2
AAA
AAG
AAAG
2
A
TG
TGAATG
4
A
G
C
T
AGT
0
Sample Output
Case 1: 1
Case 2: 4
Case 3: -1
对给出的得病的串进行自动机构造，dp[i][j]对应了走i步从根到达第j个节点的最小值。
#include
#include
#include
#include
using namespace std ; #define INF 0x3f3f3f3f struct node{ int flag , id ; node *next[4] , *fail ; } tree[2100] ; queue
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 ; }