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 0Sample 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