poj 3691 DNA repair

DNA repair

http://poj.org/problem?id=3691

Time Limit: 2000MS		Memory Limit: 65536K

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

Source

2008 Asia Hefei Regional Contest Online by USTC

 

题意：

给出n个模式串，和一个长度为m的原串，求最少修改几位，使得其中不包含任何一个模式串为子串

字母只有AGCT

 

dp[i][j]表示已经修改好了前i位u，当前在AC自动机的j节点，且前i位不包含任何一个模式串位子串的最小修改次数

不改，dp[i][j]=dp[i+1][k]

改 ,dp[i][j]=dp[i+1][k]+1

取min

如果dp[i+1][k]是单词节点就不能转移

记忆化搜索即可

#include<queue>
#include<cstdio>
#include<cstring>
using namespace std;
int n,len,root,id,ans;
char s[1001];
bool mark[51*21],v[1001][51*21];
int trie[51*21][5],tot,f[51*21],dp[1001][51*21];
queue<int>q;
struct ACautomata
{
    int get(char c)
    {
        if(c=='A') return 0;
        if(c=='G') return 1;
        if(c=='C') return 2;
        if(c=='T') return 3;
    }
    void insert()
    {
        root=1;
        len=strlen(s);
        for(int i=0;i<len;i++)
        {
            id=get(s[i]);
            if(!trie[root][id]) 
            {
                trie[root][id]=++tot;
                memset(trie[tot],0,sizeof(trie[tot]));
                mark[tot]=0;
            }
            root=trie[root][id];
        }
        mark[root]=true;
    }
    void getfail()
    {
        memset(f,0,sizeof(f));
        q.push(1);
        int now,j;
        while(!q.empty())
        {
            now=q.front(); q.pop();
            for(int i=0;i<4;i++)
            {
                if(!trie[now][i])
                {
                    trie[now][i]=trie[f[now]][i];
                    continue;
                }
                q.push(trie[now][i]);
                j=f[now];
                f[trie[now][i]]=trie[j][i];
                if(mark[trie[j][i]]) mark[trie[now][i]]=true;
            }
        }
    }
    int dfs(int l,int now)
    {
        if(l==len) return 0;
        if(v[l][now]) return dp[l][now];
        v[l][now]=true;
        if(!mark[trie[now][get(s[l+1])]]) 
          dp[l][now]=dfs(l+1,trie[now][get(s[l+1])]);
        else dp[l][now]=2000;
        for(int i=0;i<4;i++)
         if(!mark[trie[now][i]]&&i!=get(s[l+1]))  
           dp[l][now]=min(dp[l][now],dfs(l+1,trie[now][i])+1);
        return dp[l][now];
    }
};
ACautomata AC;
int main()
{
    for(int i=0;i<4;i++) trie[0][i]=1;
    int t=0;
    while(scanf("%d",&n)!=EOF)
    {
        if(!n) return 0;
        tot=1;
        memset(v,0,sizeof(v));
        memset(trie[1],0,sizeof(trie[1]));
        memset(dp,0,sizeof(dp));
        while(n--)
        {
            scanf("%s",s);
            AC.insert();
        }
        AC.getfail();
        scanf("%s",s+1);
        len=strlen(s+1);
        ans=AC.dfs(0,1);
        if(ans==2000) ans=-1;
        printf("Case %d: %d\n",++t,ans);
    }
}