http://acm.hdu.edu.cn/showproblem.php?pid=2457
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 -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
/**
hdu2457 AC自动机+DP
题目大意：给定一些模式串，对于一个字符串最少改动几个字母就能保证该字符串中不含任何一个模式串
解题思路：dp[i][j]表示长度为i以状态j为结尾改动最小值，用自动机转移一下就可以了
*/
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <queue>
#include <algorithm>
using namespace std;
const int INF=0x3f3f3f3f;
struct Trie
{
int next[1010][4],fail[1010],end[1010];
int root,L;
int newnode()
{
for(int i=0;i<4;i++)
{
next[L][i]=-1;
}
end[L++]=false;
return L-1;
}
void init()
{
L=0;
root=newnode();
}
int getch(char ch)
{
if(ch=='A')return 0;
if(ch=='C')return 1;
if(ch=='G')return 2;
return 3;
}
void insert(char buf[])
{
int len=strlen(buf);
int now=root;
for(int i=0;i<len;i++)
{
if(next[now][getch(buf[i])]==-1)
next[now][getch(buf[i])]=newnode();
now=next[now][getch(buf[i])];
}
end[now]=true;
}
void build()
{
queue<int>Q;
fail[root]=root;
for(int i=0;i<4;i++)
{
if(next[root][i]==-1)
next[root][i]=root;
else
{
fail[next[root][i]]=root;
Q.push(next[root][i]);
}
}
while(!Q.empty())
{
int now=Q.front();
Q.pop();
if(end[fail[now]])end[now]=true;
for(int i=0;i<4;i++)
{
if(next[now][i]==-1)
next[now][i]=next[fail[now]][i];
else
{
fail[next[now][i]]=next[fail[now]][i];
Q.push(next[now][i]);
}
}
}
}
int dp[1010][1010];
int solve(char *buf)
{
int len=strlen(buf);
for(int i=0;i<=len;i++)
{
for(int j=0;j<L;j++)
{
dp[i][j]=INF;
}
}
dp[0][root]=0;
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 news=next[j][k];
if(end[news])continue;
int tmp;
if(k==getch(buf[i]))tmp=dp[i][j];
else tmp=dp[i][j]+1;
dp[i+1][news]=min(dp[i+1][news],tmp);
}
}
}
}
int ans=INF;
for(int j=0;j<L;j++)
{
ans=min(ans,dp[len][j]);
}
if(ans==INF)ans=-1;
return ans;
}
}ac;
char buf[1010];
int main()
{
int n,tt=0;
while(~scanf("%d",&n))
{
if(n==0)break;
ac.init();
while(n--)
{
scanf("%s",buf);
ac.insert(buf);
}
ac.build();
scanf("%s",buf);
printf("Case %d: %d\n",++tt,ac.solve(buf));
}
return 0;
}
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 -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