Bzoj4044 [Cerc2014] Virus synthesis

Time Limit: 20 Sec Memory Limit: 128 MB
Submit: 215 Solved: 83

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

Sample Output

3
8
6
18

HINT

Source

字符串回文自动机 DP

手推一波可以发现，一个串是由它所包含的某个回文串加上其他的零散字符拼成的。

设g[S]为拼成字符串S的最小步数，有：g[S]=min{g[C]+len[S]-len[C]} （C是被S包含的一个回文串）

找回文串的过程可以用回文自动机完成，建回文自动机的时候，DP也可以顺便做完。

无脑初始化被常数卡飞，用某个点时才初始化它就过了

 1 #include<iostream>
 2 #include<algorithm>
 3 #include<cstdio>
 4 #include<cmath>
 5 #include<cstring>
 6 using namespace std;
 7 const int mxn=200010;
 8 int a[150];
 9 int T,ans,tot;
10 char s[mxn];
11 struct HWM{
12     int t[mxn][5];
13     int fa[mxn],hf[mxn],S,last,cnt;
14     int len[mxn];
15     int g[mxn];
16     void clear(int x){memset(t[x],0,sizeof t[x]);}
17     void init(){
18         S=last=cnt=1;
19         fa[0]=fa[1]=1;
20         hf[0]=hf[1]=1;
21         clear(0);clear(1);
22         g[0]=1;
23         len[0]=0;len[1]=-1;
24         return;
25     }
26     void insert(int c,int n){
27         int p=last;
28         while(s[n-len[p]-1]!=s[n])p=fa[p];
29         if(!t[p][c]){
30             int np=++cnt;
31             t[p][c]=np;
32             clear(np);
33             len[np]=len[p]+2;
34             if(p==1){
35                 fa[np]=hf[np]=0;
36             }
37             else{
38                 int k=fa[p];
39                 while(s[n-len[k]-1]!=s[n])k=fa[k];
40                 fa[np]=t[k][c];
41                 k=hf[p];
42                 while(s[n-len[k]-1]!=s[n] || (len[np]<((len[k]+2)<<1)) ){
43                     k=fa[k];
44                 }
45                 hf[np]=t[k][c];
46             }
47             if(len[np]&1){
48                 g[np]=min(len[np],g[hf[np]]+(len[np]-len[hf[np]]));
49             }
50             else{
51                 g[np]=min(g[p]+1,g[hf[np]]+(len[np]>>1)-len[hf[np]]+1);
52             }
53             ans=min(ans,tot-len[np]+g[np]);
54 //            printf("len:%d g:%d hf_len:%d ans:%d sz:%d
",len[np],g[np],g[hf[np]],ans,tot);
55         }
56         last=t[p][c];
57         return;
58     }
59 }hm;
60 int main(){
61     int i,j;
62     a['A']=1;a['G']=2;a['C']=3;a['T']=4;
63     scanf("%d",&T);
64     while(T--){
65         scanf("%s",s+1);
66         hm.init();
67         int n=strlen(s+1);tot=n;
68         ans=n;
69         for(i=1;i<=n;i++)
70             hm.insert(a[s[i]],i);
71         printf("%d
",ans);
72     }
73     return 0;
74 }