【POJ3691】 DNA repair （AC自动机+DP）

DNA repair
Time Limit: 2000MS Memory Limit: 65536KB 64bit IO Format: %I64d & %I64u

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
 
【题意】
　　已知一个DNA串和一些病毒DNA序列，求出最少改变DNA串中多少个字符，能使得串中不包含任意一个病毒序列。
 

【分析】

　　建AC自动机然后DP，不要走到有标记的点即可。

 

 代码如下：

  1 #include<cstdio>
  2 #include<cstdlib>
  3 #include<cstring>
  4 #include<iostream>
  5 #include<algorithm>
  6 #include<queue>
  7 using namespace std;
  8 #define Maxn 3010
  9 #define Maxl 1010
 10 #define INF 0xfffffff
 11 
 12 int n;
 13 char s[20];
 14 char ss[Maxl];
 15 
 16 struct node
 17 {
 18     int cnt,fail;
 19     bool mark;
 20     int son[5];
 21 }t[Maxn];int tot;//=0;
 22 
 23 void upd(int x)
 24 {
 25     t[x].cnt=1;t[x].mark=0;
 26     memset(t[x].son,0,sizeof(t[x].son));
 27 }
 28 
 29 int mymin(int x,int y) {return x<y?x:y;}
 30 
 31 void read_trie()
 32 {
 33     scanf("%s",s+1);
 34     int len=strlen(s+1);
 35     int now=0;
 36     for(int i=1;i<=len;i++)
 37     {
 38         int ind=s[i]-'A'+1;
 39         if(ind==3) ind=2;
 40         else if(ind==7) ind=3;
 41         else if(ind==20) ind=4;
 42         if(!t[now].son[ind]) 
 43         {
 44             t[now].son[ind]=++tot;
 45             upd(tot);
 46         }
 47         now=t[now].son[ind];
 48         if(i==len) t[now].mark=1;
 49     }
 50 }
 51 
 52 queue<int > q;
 53 // bool inq[Maxn];
 54 void build_AC()
 55 {
 56     while(!q.empty()) q.pop();
 57     q.push(0);//inq[0]=1;
 58     while(!q.empty())
 59     {
 60         int x=q.front();q.pop();
 61         for(int i=1;i<=4;i++) 
 62         {
 63             if(t[x].son[i])
 64             {
 65                 t[t[x].son[i]].fail=x?t[t[x].fail].son[i]:0;
 66                 q.push(t[x].son[i]);
 67             }
 68             else t[x].son[i]=t[t[x].fail].son[i];
 69             if(t[t[x].fail].mark) t[x].mark=1;
 70         }
 71     }
 72 }
 73 
 74 int f[Maxn][Maxl];
 75 void dp()
 76 {
 77     scanf("%s",ss+1);
 78     int len=strlen(ss+1);
 79     for(int i=1;i<=len;i++)
 80     {
 81         if(ss[i]=='C') ss[i]='B';
 82         else if(ss[i]=='G') ss[i]='C';
 83         else if(ss[i]=='T') ss[i]='D';
 84     }
 85     memset(f,63,sizeof(f));
 86     f[0][0]=0;
 87     for(int i=1;i<=len;i++)
 88      for(int j=0;j<=tot;j++) if(f[i-1][j]<INF)
 89      {
 90          for(int k=1;k<=4;k++) if(!t[t[j].son[k]].mark)
 91          {
 92              if(ss[i]-'A'+1==k) 
 93               f[i][t[j].son[k]]=mymin(f[i][t[j].son[k]],f[i-1][j]);
 94              else f[i][t[j].son[k]]=mymin(f[i][t[j].son[k]],f[i-1][j]+1);
 95          }
 96      }
 97     int ans=INF;
 98     for(int i=0;i<=tot;i++) ans=mymin(f[len][i],ans);
 99     if(ans<=len) printf("%d\n",ans);
100     else printf("-1\n");
101 }
102 
103 void init()
104 {
105     tot=0;
106     upd(0);
107     for(int i=1;i<=n;i++)
108     {
109         read_trie();
110     }
111     build_AC();
112 }
113 
114 int main()
115 {
116     int kase=0;
117     while(1)
118     {
119         scanf("%d",&n);
120         if(n==0) break;
121         init();
122         printf("Case %d: ",++kase);
123         dp();
124     }
125     return 0;
126 }
[POJ3691]

 

2016-07-11 10:06:17

 

 

posted @ 2016-07-11 10:03  konjak魔芋  阅读(582)  评论(0)  编辑  收藏  举报