1605 - Gene recombination
Time Limit: 2s Memory Limit: 64MB
Submissions: 264 Solved: 46
- DESCRIPTION
- As a gene engineer of a gene engineering project, Enigma encountered a puzzle about gene recombination. It is well known that a gene can be considered as a sequence, consisting of four nucleotides, which are simply denoted by four letters, A, C, G, and T.
Enigma has gotten a gene, such as "ATCC". And he wants to reorder this gene to make a new one, like "CTCA". He can use two types of operations: (1) exchange the first two letters, or (2) move the first letter to the end of the gene. For example, "ATCC" can be changed to "TCCA" by operation (2), and then "TCCA" can be changed to "CTCA" by operation (1). Your task is to make a program to help Enigma to find out the minimum number of operations to reorder the gene. - INPUT
- The input contains several test cases. The first line of a test case contains one integer N indicating the length of the gene (1<=N<=12). The second line contains a string indicating the initial gene. The third line contains another string which Enigma wants.
Note that the two strings have the same number for each kind of letter. - OUTPUT
- For each test case, output the minimum number of operations.
- SAMPLE INPUT
-
4 ATCC CTCA 4 ATCG GCTA 4 ATCG TAGC
- SAMPLE OUTPUT
-
2 4 6
- 题目大意：给两个长度不超过十二的字符串S1，S2（由A、C、G、T组成），有两种操作:1.交换第一个跟第二个字母。2.把第一个移到字符串的末尾。
- 求最少的操作次数S1==S2。
- 分析：用BFS深搜加字典树剪枝。
1 #include<iostream> 2 #include<cstdio> 3 #include<string> 4 #include<cstring> 5 #include<queue> 6 using namespace std; 7 8 const int maxn=5000000; 9 string s1,s2; 10 struct Trie //字典树查询字符串 11 { 12 int ch[maxn][4]; 13 int sz; //结点总数 14 void clear(){sz=1;memset(ch[0],0,sizeof(ch[0]));} 15 int Num(char ch) 16 { 17 if(ch=='A') return 0; 18 if(ch=='C') return 1; 19 if(ch=='G') return 2; 20 if(ch=='T') return 3; 21 } 22 void insert(string s) 23 { 24 int len=s.size(),u=0; 25 for(int i=0;i<len;i++) 26 { 27 int c=Num(s[i]); 28 if(!ch[u][c]) 29 { 30 memset(ch[sz],0,sizeof(ch[sz])); 31 ch[u][c]=sz++; 32 } 33 u=ch[u][c]; 34 } 35 } 36 int Find(string s) 37 { 38 int u = 0,len=s.size(); 39 for(int i = 0; i < len; i++) 40 { 41 int c = Num(s[i]); 42 if(!ch[u][c]) return 0; 43 u = ch[u][c]; 44 } 45 return 1; 46 } 47 }trie; 48 struct point //队列数据 49 { 50 string s; 51 int step; 52 }; 53 void swap(char &a,char &b) 54 { 55 char t=a; 56 a=b; 57 b=t; 58 } 59 string change1(string s) 60 { 61 swap(s[0],s[1]); 62 return s; 63 } 64 string change2(string s) 65 { 66 char ch=s[0]; 67 s.erase(s.begin()); 68 s+=ch; 69 return s; 70 } 71 void solve() 72 { 73 trie.clear(); 74 queue<point> Q; 75 point p,t; 76 t.s=s1;t.step=0; 77 trie.insert(t.s); 78 Q.push(t); 79 while(!Q.empty()) 80 { 81 t=Q.front();Q.pop(); 82 p.step=t.step+1; 83 p.s=change1(t.s); 84 if(!trie.Find(p.s)) 85 { 86 if(p.s==s2) 87 { 88 printf("%d ",p.step); 89 return ; 90 } 91 Q.push(p); 92 trie.insert(p.s); 93 } 94 p.s=change2(t.s); 95 if(!trie.Find(p.s)) 96 { 97 if(p.s==s2) 98 { 99 printf("%d ",p.step); 100 return ; 101 } 102 Q.push(p); 103 trie.insert(p.s); 104 } 105 } 106 return ; 107 } 108 int main() 109 { 110 int n; 111 while(~scanf("%d",&n)) 112 { 113 cin>>s1>>s2; 114 if(s1==s2) 115 { 116 printf("0 "); 117 continue; 118 } 119 solve(); 120 } 121 return 0; 122 }