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.
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.
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.
2008 Asia Hefei Regional Contest Online by USTC有个题解写的挺好的
看这里
http://www.cnblogs.com/woodfish1988/archive/2008/10/03/1303492.html
就是用trie图来表示状态及状态之间的转移
code
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <ctime>
#include <cassert>
#include <iostream>
#include <sstream>
#include <fstream>
#include <map>
#include <set>
#include <vector>
#include <queue>
#include <algorithm>
#include <iomanip>
#define maxn 1105
#define inf 0x7fffffff
using namespace std;
struct node
{
int next[4];
int fail,count;
void init()
{
memset(next,-1,sizeof(next));
fail=-1;
count=0;
}
} s[maxn];
int f[maxn][maxn];
int hash[256];
int sind,times,n;
int q[maxn],head,tail;
void cas_init()
{
s[0].init();
sind=1;
}
void ins(char str[])
{
int i,j,len,ind;
len=strlen(str);
ind=0;
for(i=0; i<len; i++)
{
j=hash[str[i]];
if(s[ind].next[j]==-1)
{
s[sind].init();
s[ind].next[j]=sind++;
}
ind=s[ind].next[j];
}
s[ind].count++;
}
void make_fail()
{
int u,i,p,son;
head=0;
tail=1;
q[tail]=0;
while(head<tail)
{
head++;
u=q[head];
for(i=0; i<4; i++)
{
if(s[u].next[i]!=-1)
{
p=s[u].fail;
son=s[u].next[i];
while(p!=-1&&s[p].next[i]==-1) p=s[p].fail;
if(u==0) s[son].fail=0;
else s[son].fail=s[p].next[i];
if(s[s[son].fail].count) s[son].count=1;
q[++tail]=son;
}
else
{
p=s[u].fail;
while(p!=-1&&s[p].next[i]==-1)
p=s[p].fail;
if(u==0) s[u].next[i]=0;
else s[u].next[i]=s[p].next[i];
}
}
}
}
int min(int a,int b)
{
if(a==-1) return b;
if(b==-1) return a;
return a<b?a:b;
}
void getans(char str[])
{
int v,i,j,k;
int ans=inf;
int len;
len=strlen(str);
memset(f,-1,sizeof(f));
f[0][0]=0;
for(i=1; i<=len; i++)
{
for(j=0; j<sind; j++)
if(f[i-1][j]!=-1)
{
for(k=0; k<4; k++)
{
v=s[j].next[k];
if(s[v].count==0)
{
f[i][v]=min(f[i][v],f[i-1][j]+(k!=hash[str[i-1]]));
}
}
}
}
//printf("%d\n",i);
ans=-1;
for(i=0; i<sind; i++)
{
if(s[i].count==0)
{
ans=min(ans,f[len][i]);
}
}
printf("Case %d: %d\n",++times,ans);
}
int main()
{
char str[50],str1[1005];
hash['A']=0;
hash['T']=1;
hash['G']=2;
hash['C']=3;
times=0;
while(scanf("%d",&n)!=EOF&&n!=0)
{
// printf("%d\n",n);
cas_init();
for(int i=1; i<=n; i++)
{
scanf("%s",str);
ins(str);
}
make_fail();
scanf("%s",str1);
getans(str1);
}
return 0;
}