Total Submission(s): 1930 Accepted Submission(s): 1046
Problem Description
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.
You are to help the biologists to repair a DNA by changing least number of characters.
Input
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.
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.
Output
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.
number of characters which need to be changed. If it's impossible to repair the given DNA, print -1.
Sample Input
2
AAA
AAG
AAAG
2
A
TG
TGAATG
4
A
G
C
T
AGT
0
Sample Output
Case 1: 1
Case 2: 4
Case 3: -1
题意：给你n个模板串和一个字符串，让你对字符串里的字符进行修改，使得修改后不含模板串中的任何一个，求最小的修改次数。
思路：先构造trie图，其实trie图里的next数组就是状态转移过程了，我们只要定义dp[i][j]表示长度为i，到节点j所要修改的最少次数。那么if(t==idx(s[i]) )dp[i+1][jj ]=min(dp[i+1][jj],dp[i][j] );else{dp[i+1][jj]=min(dp[i+1 ][jj],dp[i][j]+1 );}
#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<string>
#include<algorithm>
using namespace std;
typedef long long ll;
#define inf 99999999
#define pi acos(-1.0)
#define maxnode 1200
int dp[maxnode][maxnode];
char s[55],str[1006];
int tot=0;
struct trie{
int sz,root,val[maxnode],next[maxnode][6],fail[maxnode];
int q[1111111];
void init(){
int i;
sz=root=0;
val[0]=0;
for(i=0;i<4;i++){
next[root][i]=-1;
}
}
int idx(char c){
if(c=='A')return 0;
if(c=='G')return 1;
if(c=='C')return 2;
if(c=='T')return 3;
}
void charu(char *s){
int i,j,u=0;
int len=strlen(s);
for(i=0;i<len;i++){
int c=idx(s[i]);
if(next[u][c]==-1){
sz++;
val[sz]=0;
next[u][c]=sz;
u=next[u][c];
for(j=0;j<4;j++){
next[u][j]=-1;
}
}
else{
u=next[u][c];
}
}
val[u]=1;
}
void build(){
int i,j;
int front,rear;
front=1;rear=0;
for(i=0;i<4;i++){
if(next[root][i]==-1 ){
next[root][i]=root;
}
else{
fail[next[root][i] ]=root;
rear++;
q[rear]=next[root][i];
}
}
while(front<=rear){
int x=q[front];
if(val[fail[x] ])val[x]=1;
front++;
for(i=0;i<4;i++){
if(next[x][i]==-1){
next[x][i]=next[fail[x] ][i];
}
else{
fail[next[x][i] ]=next[fail[x] ][i];
rear++;
q[rear]=next[x][i];
}
}
}
}
void solve(char *s)
{
int i,j,len,t,jj;
len=strlen(s);
for(i=0;i<=len;i++){
for(j=0;j<=sz;j++){
dp[i][j]=inf;
}
}
dp[0][0]=0;
for(i=0;i<=len-1;i++){
for(j=0;j<=sz;j++){
if(dp[i][j]==inf)continue;
for(t=0;t<4;t++){
jj=next[j][t];
if(val[jj ])continue;
if(t==idx(s[i]) )dp[i+1][jj ]=min(dp[i+1][jj],dp[i][j] );
else{
dp[i+1][jj]=min(dp[i+1 ][jj],dp[i][j]+1 );
}
}
}
}
int minx=inf;
for(j=0;j<=sz;j++){
if(val[j])continue;
minx=min(minx,dp[len][j]);
}
printf("Case %d: ",++tot);
if(minx>1000000)printf("-1
");
else printf("%d
",minx);
}
}ac;
int main()
{
int n,m,i,j;
while(scanf("%d",&n)!=EOF && n!=0)
{
ac.init();
for(i=1;i<=n;i++){
scanf("%s",s);
ac.charu(s);
}
ac.build();
scanf("%s",str);
ac.solve(str);
}
return 0;
}