Problem 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 ⑴.
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: ⑴
Source
2008 Asia Hefei Regional Contest Online by USTC
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設(shè)每個(gè)模式串結(jié)尾所在的節(jié)點(diǎn)為危險(xiǎn)節(jié)點(diǎn),明顯在AC自動(dòng)機(jī)上,如果1個(gè)節(jié)點(diǎn)的fail指針指向的那個(gè)節(jié)點(diǎn)是危險(xiǎn)節(jié)點(diǎn),那末這個(gè)節(jié)點(diǎn)是危險(xiǎn)節(jié)點(diǎn),由于它們是后綴關(guān)系
設(shè)dp[i][j]表示長度為i,在節(jié)點(diǎn)j時(shí),且不包括任何危險(xiǎn)節(jié)點(diǎn),所需要改變的最少的字符數(shù)
/*************************************************************************
> File Name: hdu2457.cpp
> Author: ALex
> Mail: zchao1995@gmail.com
> Created Time: 2015年03月05日 星期4 14時(shí)29分23秒
************************************************************************/
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <vector>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const double pi = acos(-1);
const int inf = 0x3f3f3f3f;
const double eps = 1e⑴5;
typedef long long LL;
typedef pair <int, int> PLL;
const int MAX_NODE = 2010;
const int CHILD_NUM = 4;
int dp[1010][MAX_NODE];
struct AC_Automation
{
int next[MAX_NODE][CHILD_NUM];
int fail[MAX_NODE];
int end[MAX_NODE];
int root, L;
int ID (char c)
{
if (c == 'A')
{
return 0;
}
if (c == 'G')
{
return 1;
}
if (c == 'C')
{
return 2;
}
if (c == 'T')
{
return 3;
}
}
int newnode ()
{
for (int i = 0; i < CHILD_NUM; ++i)
{
next[L][i] = -1;
}
end[L++] = 0;
return L - 1;
}
void init ()
{
L = 0;
root = newnode ();
}
void Build_Trie (char buf[])
{
int now = root;
int len = strlen (buf);
for (int i = 0; i < len; ++i)
{
if (next[now][ID(buf[i])] == -1)
{
next[now][ID(buf[i])] = newnode();
}
now = next[now][ID(buf[i])];
}
end[now] = 1;
}
void Build_AC ()
{
queue <int> qu;
fail[root] = root;
for (int i = 0; i < CHILD_NUM; ++i)
{
if (next[root][i] == -1)
{
next[root][i] = root;
}
else
{
fail[next[root][i]] = root;
qu.push (next[root][i]);
}
}
while (!qu.empty())
{
int now = qu.front();
qu.pop();
if (end[fail[now]])
{
end[now] = 1;
}
for (int i = 0; i < CHILD_NUM; ++i)
{
if (next[now][i] == -1)
{
next[now][i] = next[fail[now]][i];
}
else
{
fail[next[now][i]] = next[fail[now]][i];
qu.push (next[now][i]);
}
}
}
}
void solve(char buf[])
{
memset (dp, inf, sizeof(dp));
dp[0][0] = 0;
int len = strlen (buf);
for (int i = 0; i < len; ++i)
{
for (int j = 0; j < L; ++j)
{
if (dp[i][j] < inf)
{
for (int k = 0; k < 4; ++k)
{
int now = next[j][k];
if (end[now])
{
continue;
}
int use = dp[i][j];
if (k != ID(buf[i]))
{
++use;
}
dp[i + 1][now] = min (dp[i + 1][now], use);
}
}
}
}
int ans = inf;
for (int i = 0; i < L; ++i)
{
ans = min (ans, dp[len][i]);
}
printf("%d
", ans >= inf ? -1 : ans);
}
}AC;
char buf[1010];
int main ()
{
int n;
int icase = 1;
while (~scanf("%d", &n), n)
{
printf("Case %d: ", icase++);
AC.init();
for (int i = 1; i <= n; ++i)
{
scanf("%s", buf);
AC.Build_Trie (buf);
}
AC.Build_AC();
scanf("%s", buf);
AC.solve (buf);
}
return 0;
}
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