hdu 2457（AC自动机+DP）

DNA repair

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

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

 

解题思路：题目的模型有点像是AC自动机的模型，只是AC自动机是查询有多少个子串在文本串中出现过，不过我们可以利用一下这个性质。

做法：

1.对n个短字符串建一颗字典树

2.构造AC自动机，注意的是，在构建失败指针的时候，如果一个节点的失败指针所指的节点是那个病毒串的话，也就是说是dangerous的话，那么这个节点也是dangerous的

3.设dp[i][j]表示长度为i的字符串到达j节点时所需改变最小字符个数，然后根据AC自动机进行状态转移

感觉做AC自动机+DP的问题，有一个状态一定是Trie树中节点的转移。。。

#include<iostream>#include<cstdio>#include<cstring>#include<cmath>#include<algorithm>#define N 100005#define MOD 100000#define inf 1<<29#define LL long longusing namespace std;struct Trie{    Trie *next[4];    Trie *fail;    int kind,isword;};Trie *que[N],s[N];int idx;int id(char ch){    if(ch=='A') return 0;    else if(ch=='T') return 1;    else if(ch=='C') return 2;    return 3;}Trie *NewNode(){    Trie *tmp=&s[idx];    for(int i=0;i<4;i++) tmp->next[i]=NULL;    tmp->isword=0;    tmp->kind=idx++;    tmp->fail=NULL;    return tmp;}void Insert(Trie *root,char *s,int len){    Trie *p=root;    for(int i=0;i<len;i++){        if(p->next[id(s[i])]==NULL)            p->next[id(s[i])]=NewNode();        p=p->next[id(s[i])];    }    p->isword=1;}void Bulid_Fail(Trie *root){    int head=0,tail=0;    que[tail++]=root;    root->fail=NULL;    while(head<tail){        Trie *tmp=que[head++];        for(int i=0;i<4;i++){            if(tmp->next[i]){                if(tmp==root) tmp->next[i]->fail=root;                else{                    Trie *p=tmp->fail;                    while(p!=NULL){                        if(p->next[i]){                           tmp->next[i]->fail=p->next[i];                           break;                        }                        p=p->fail;                    }                    if(p==NULL) tmp->next[i]->fail=root;                }                if(tmp->next[i]->fail->isword) tmp->next[i]->isword=1;                que[tail++]=tmp->next[i];            }            else if(tmp==root) tmp->next[i]=root;            else tmp->next[i]=tmp->fail->next[i];        }    }}int dp[1005][2005];int slove(char *str,int len){    for(int i=0;i<=len;i++) for(int j=0;j<idx;j++) dp[i][j]=inf;    dp[0][0]=0;    for(int i=1;i<=len;i++){        for(int j=0;j<idx;j++){            if(s[j].isword) continue;            if(dp[i-1][j]==inf) continue;            for(int k=0;k<4;k++){                int r=s[j].next[k]->kind;                if(s[r].isword) continue;                dp[i][r]=min(dp[i][r],dp[i-1][j]+(id(str[i-1])!=k));            }        }    }    int ans=inf;    for(int i=0;i<idx;i++) ans=min(ans,dp[len][i]);    return ans==inf?-1:ans;}char str[1005];int main(){    int n,cas=0;    while(scanf("%d",&n)!=EOF&&n){        idx=0;        Trie *root=NewNode();        for(int i=0;i<n;i++){            scanf("%s",str);            Insert(root,str,strlen(str));        }        Bulid_Fail(root);        scanf("%s",str);        printf("Case %d: %d\n",++cas,slove(str,strlen(str)));    }    return 0;}