AC自动机+dp。问改变多少个字符能让目标串不含病毒串。即走过多少步不经过病毒串终点。又是同样的问题。
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<queue>
using namespace std;
#define REP(i,s,t) for(int i=s;i<=t;i++)
#define dwn(i,s,t) for(int i=s;i>=t;i--)
#define clr(x,c) memset(x,c,sizeof(x))
const int nmax=1005;
const int inf=0x7f7f7f7f;
int ch[nmax][4],fail[nmax],pt=0,dp[nmax][nmax];
bool F[nmax];
char s[nmax];
int id(char c){
if(c=='A') return 0;
if(c=='C') return 1;
if(c=='G') return 2;
else return 3;
}
void insert(){
int t=0,len=strlen(s);
REP(i,0,len-1) {
if(!ch[t][id(s[i])]) ch[t][id(s[i])]=++pt;
t=ch[t][id(s[i])];
}
F[t]=true;
}
queue<int>q;
void getfail(){
q.push(0);fail[0]=0;
while(!q.empty()){
int x=q.front();q.pop();
REP(i,0,3){
if(ch[x][i]) q.push(ch[x][i]),fail[ch[x][i]]=x==0?0:ch[fail[x]][i];
else ch[x][i]=x==0?0:ch[fail[x]][i];
}
F[x]|=F[fail[x]];
}
}
void work(int x){
clr(dp,0x7f);dp[0][0]=0;
scanf("%s",s);int len=strlen(s);
REP(i,0,len-1) REP(j,0,pt) if(dp[i][j]!=inf){
REP(k,0,3){
int tmp=ch[j][k];if(F[tmp]) continue;
int temp=(k==id(s[i]))?dp[i][j]:dp[i][j]+1;
dp[i+1][tmp]=min(dp[i+1][tmp],temp);
}
}
int ans=inf;
REP(i,0,pt) ans=min(ans,dp[len][i]);
if(ans==inf) printf("Case %d: -1\n",x);
else printf("Case %d: %d\n",x,ans);
}
int main(){
int n,cur=0;
while(scanf("%d",&n)!=EOF&&n){
clr(fail,0);clr(ch,0);clr(F,false);pt=0;
REP(i,1,n) scanf("%s",s),insert();
getfail();work(++cur);
}
return 0;
}
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DNA repairTime 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 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 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. 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 Source
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