Description
Viruses are usually bad for your health. How about fighting them with... other viruses? In
this problem, you need to find out how to synthesize such good viruses.
We have prepared for you a set of strings of the letters A, G, T and C. They correspond to the
DNA nucleotide sequences of viruses that we want to svnthesize, using the following operations:
* Adding a nucleotide either to the beginning or the end of the existing sequence
* Replicating the sequence, reversing the copied piece, and gluing it either to the beginmng or
to the end of the original (so that e.g., AGTC can become AGTCCTGA or CTGAAGTC).
We're concerned about efficiency, since we have very many such sequences, some of them verv
long. Find a wav to svnthesize them in a mmimum number of operations.
你要用ATGC四个字母用两种操作拼出给定的串：
1.将其中一个字符放在已有串开头或者结尾
2.将已有串复制，然后reverse，再接在已有串的头部或者尾部
一开始已有串为空。求最少操作次数。
len<=100000
Input
The first line of input contains the number of test cases T. The descriptions of the test cases
follow:
Each test case consists of a single line containing a non-empty string. The string uses only
the capital letters A, C, G and T and is not longer than 100 000 characters.
Output
For each test case, output a single line containing the minimum total number of operations
necessary to construct the given sequence.
Sample Input
4
AAAA
AGCTTGCA
AAGGGGAAGGGGAA
AAACAGTCCTGACAAAAAAAAAAAAC
AAAA
AGCTTGCA
AAGGGGAAGGGGAA
AAACAGTCCTGACAAAAAAAAAAAAC
Sample Output
3
8
6
18
8
6
18
解题思路：
有这样的性质：
1.只能形成1个大回文串。
2.只能生成偶回文串。 很有动归的思想嘛。 回文树上一个节点代表一个回文串，dp[i]表示在i节点的回文串最小生成代价：
其中mid为长度小于len[i]/2的最长后缀回文串所在节点，可以倍增跳也可以从fa的mid开始跳。
那么答案就是：
就愉快地结束了。
代码：
1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 struct pant{ 5 int tranc[4]; 6 int len; 7 int pre; 8 int dp; 9 int mid; 10 }h[1000000],stpant; 11 int a[100001]; 12 char tmp[100001]; 13 int siz; 14 int fin; 15 int len; 16 int trans(char t) 17 { 18 if(t=='A') 19 return 0; 20 if(t=='G') 21 return 1; 22 if(t=='C') 23 return 2; 24 return 3; 25 } 26 void Res(void) 27 { 28 memset(a,0x3f,sizeof(a)); 29 a[0]=-1; 30 h[1]=h[0]=stpant; 31 h[1].pre=h[0].pre=1; 32 h[1].len=-1; 33 h[0].dp=1; 34 siz=1; 35 fin=0; 36 return ; 37 } 38 bool mis(int i,int lsp) 39 { 40 return a[i]!=a[i-h[lsp].len-1]; 41 } 42 void Insert(int i) 43 { 44 int nwp,lsp,mac; 45 lsp=fin; 46 int c=a[i]; 47 while(mis(i,lsp)) 48 lsp=h[lsp].pre; 49 if(!h[lsp].tranc[c]) 50 { 51 nwp=++siz; 52 mac=h[lsp].pre; 53 h[nwp]=stpant; 54 h[nwp].len=h[nwp].dp=h[lsp].len+2; 55 while(mis(i,mac)) 56 mac=h[mac].pre; 57 h[nwp].pre=h[mac].tranc[c]; 58 h[lsp].tranc[c]=nwp; 59 60 61 if(h[nwp].len<=2) 62 h[nwp].mid=h[nwp].pre; 63 else{ 64 mac=h[lsp].mid; 65 while(mis(i,mac)||h[mac].len*2+4>h[nwp].len) 66 mac=h[mac].pre; 67 h[nwp].mid=h[mac].tranc[c]; 68 } 69 if(h[nwp].len%2==0) 70 { 71 h[nwp].dp=std::min(h[lsp].dp+1,h[nwp].len/2-h[h[nwp].mid].len+h[h[nwp].mid].dp+1); 72 } 73 } 74 fin=h[lsp].tranc[c]; 75 return ; 76 } 77 int main() 78 { 79 int T=0; 80 scanf("%d",&T); 81 while(T--) 82 { 83 Res(); 84 scanf("%s",tmp+1); 85 int ans; 86 len=strlen(tmp+1); 87 ans=len; 88 for(int i=1;i<=len;i++) 89 a[i]=trans(tmp[i]); 90 for(int i=1;i<=len;i++) 91 Insert(i); 92 for(int i=2;i<=siz;i++) 93 { 94 if(h[i].len&1^1) 95 { 96 ans=std::min(ans,len-h[i].len+h[i].dp); 97 98 }//printf("%d ",h[i].dp); 99 } 100 //return 0; 101 printf("%d ",ans); 102 } 103 return 0; 104 }