hdu2457 DNA repair ac机+dp

DNA repair

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 1201    Accepted Submission(s): 649

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

2AAAAAGAAAG    2ATGTGAATG4AGCTAGT0

 

Sample Output

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

 

Source

2008 Asia Hefei Regional Contest Online by USTC

 

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题意：给定n个仅由A、G、C、T构成的非法串，又给定一个仅由A、G、C、T构成的字符串s，问最少需要多少步使得s不包含非法串，如若不能输出-1。

思路：先将A、C、G、T编号，并将n个非法串建ac自动机,设dp[i][j]表示长度为i的串以ac上j结点为末尾所需修改的最少步数，设p=trie[j].next[k]->id, 那么dp[i+1][p]=min(dp[i][j]+(k!=idx(str[i])),dp[i+1][p])，转移的前提是p,j结点必须均为合法结点，即该结点的cnt不为1。详见代码：

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int sigma_size=4;const int MAXN=1000+100;const int inf=0x3fffffff;int n,head,tail,sz;int dp[MAXN][MAXN];char s[MAXN];struct node{    int cnt,id;    node *next[sigma_size],*fail;}trie[MAXN],*root,*que[MAXN];struct AC{    node *createnode()    {        for(int k=0;k<sigma_size;k++)            trie[sz].next[k]=NULL;        trie[sz].fail=NULL;        trie[sz].cnt=0,trie[sz].id=sz;        return &trie[sz++];    }    void init()    {        sz=0;        head=tail=0;        root=createnode();    }    int idx(char c)    {        if(c=='A') return 0;        if(c=='G') return 1;        if(c=='C') return 2;        return 3;    }    void insert(char *str)    {        node *p=root;        int len=strlen(str);        for(int i=0;i<len;i++)        {            int k=idx(str[i]);            if(p->next[k]==NULL)                p->next[k]=createnode();            p=p->next[k];        }        p->cnt=1;    }    void get_fail()    {        que[tail++]=root;        while(head<tail)        {            node *p=que[head++];            for(int k=0;k<sigma_size;k++)                if(p->next[k])                {                    if(p==root)                        p->next[k]->fail=root;                    else                        p->next[k]->fail=p->fail->next[k];                    if(p->next[k]->fail->cnt)                        p->next[k]->cnt=1;                    que[tail++]=p->next[k];                }                else                {                    if(p==root)                        p->next[k]=root;                    else                        p->next[k]=p->fail->next[k];                }        }    }    int query(char *str)    {        int len=strlen(str);        for(int i=0;i<=len;i++)            for(int j=0;j<sz;j++)                dp[i][j]=inf;        dp[0][0]=0;        for(int i=0;i<len;i++)            for(int j=0;j<sz;j++)            {                if(trie[j].cnt)                    continue;                for(int k=0;k<sigma_size;k++)                {                    int p=trie[j].next[k]->id;                    if(trie[p].cnt)                        continue;                    dp[i+1][p]=min(dp[i+1][p],dp[i][j]+(k!=idx(str[i])));                }            }        int ans=inf;        for(int i=0;i<sz;i++)            ans=min(ans,dp[len][i]);        if(ans==inf)            ans=-1;        return ans;    }}ac;int main(){    freopen("text.txt","r",stdin);    int kase=0;    while(~scanf("%d",&n)&& n)    {        kase++;        ac.init();        for(int i=0;i<n;i++)        {            scanf("%s",s);            ac.insert(s);        }        ac.get_fail();        scanf("%s",s);        printf("Case %d: %d\n",kase,ac.query(s));    }    return 0;}