Viruses are usually bad for your health. How about ghting them with… other viruses? In this problem,
you need to nd 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 synthesize, 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 beginning 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 very long.
Find a way to synthesize them in a minimum number of operations.
Input
The rst 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
参考大牛的博客
http://www.cnblogs.com/clrs97/p/4700658.html
#include <iostream>
#include <string.h>
#include <algorithm>
#include <math.h>
#include <stdio.h>
#include <stdlib.h>
#include <map>using namespace std;
typedef long long int LL;
const int maxn=1e5+5;
map<char,int> m;
int n;
char str[maxn];
int f[maxn];
struct Tree
{
int next[maxn][4];
int fail[maxn];
int train[maxn];
int len[maxn];
int num[maxn];
int cnt[maxn];
int s[maxn];
int last;
int p;
int n;
int new_node(int x)
{
memset(next[p],0,sizeof(next[p]));
len[p]=x;
return p++;
}
void init()
{
p=0;
new_node(0);
new_node(-1);
last=0;n=0;
s[0]=-1;
fail[0]=1;
}
int get_fail(int x)
{
while(s[n-len[x]-1]!=s[n])
x=fail[x];
return x;
}
int get_fail2(int x,int pos)
{
while(s[n-len[x]-1]!=s[n]||(len[x]+2)*2>len[pos])
x=fail[x];
return x;
}
void add(char xx)
{
int x=m[xx];
s[++n]=x;
int cur=get_fail(last);
if(!(last=next[cur][x]))
{
int now=new_node(len[cur]+2);
fail[now]=next[get_fail(fail[cur])][x];
if(len[now]<=2)
train[now]=fail[now];
else
train[now]=next[get_fail2(train[cur],now)][x]; next[cur][x]=now;
last=now;
}
}
}tree;
int ans;
int q[maxn];
int rear,head;
int main()
{
int t;
scanf("%d",&t);
m['A']=0;m['C']=1;m['G']=2;m['T']=3;
while(t--)
{
scanf("%s",str);
int len=strlen(str);
tree.init();
ans=len;
for(int i=0;i<len;i++)
{
tree.add(str[i]);
}
memset(f,0,sizeof(f));
for(int i=2;i<tree.p;i++)
{
if(tree.len[i]&1)
f[i]=tree.len[i];
}
f[0]=1;
q[0]=0;
head=0;rear=1;
int x,y;
while(head<rear)
{
x=q[head++];
for(int i=0;i<4;i++)
{
if(tree.next[x][i])
{
y=tree.next[x][i];
f[y]=f[x]+1;
f[y]=min(f[y],tree.len[y]/2-tree.len[tree.train[y]]+f[tree.train[y]]+1);
ans=min(ans,len-tree.len[y]+f[y]);
q[rear++]=y;
} }
}
printf("%d\n",ans);
}
return 0;
}
