| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 6758 | Accepted: 3133 |
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.
Input
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
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
题意：给出n个模式串，和一个长度为m的原串，求最少修改几位，使得其中不包含任何一个模式串为子串
#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> using namespace std; const int N=1005,INF=1e9; inline int read(){ char c=getchar();int x=0,f=1; while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();} while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();} return x*f; } int n,m,mp[N]; char s[N]; struct node{ int ch[5],fail,val; }t[N]; int sz; void ins(char s[]){ int u=0,n=strlen(s+1); for(int i=1;i<=n;i++){ int c=mp[s[i]]; if(!t[u].ch[c]) t[u].ch[c]=++sz; u=t[u].ch[c]; } t[u].val=1; } int q[N],head,tail; void getAC(){ head=tail=1; for(int i=1;i<=4;i++) if(t[0].ch[i]) q[tail++]=t[0].ch[i]; while(head!=tail){ int u=q[head++]; t[u].val|=t[t[u].fail].val; for(int i=1;i<=4;i++){ int &v=t[u].ch[i]; if(!v) v=t[t[u].fail].ch[i]; else{ t[v].fail=t[t[u].fail].ch[i]; q[tail++]=v; } } } } int f[N][N]; void dp(){ memset(f,127,sizeof(f)); f[0][0]=0; for(int i=0;i<m;i++) for(int j=0;j<=sz;j++) if(f[i][j]<INF){ int k=t[j].ch[mp[s[i+1]]]; if(!t[k].val) f[i+1][k]=min(f[i+1][k],f[i][j]); for(int k=1;k<=4;k++) if(!t[t[j].ch[k]].val) f[i+1][t[j].ch[k]]=min(f[i+1][t[j].ch[k]],f[i][j]+1); } //for(int i=1;i<=m;i++) // for(int j=0;j<=sz;j++) printf("hi %d %d %d ",i,j,f[i][j]); int ans=INF; for(int i=0;i<=sz;i++) ans=min(ans,f[m][i]); if(ans>=INF) puts("-1"); else printf("%d ",ans); } int main(){ freopen("in","r",stdin); int cas=0; mp['A']=1;mp['G']=2;mp['C']=3;mp['T']=4; while((n=read())){ printf("Case %d: ",++cas); memset(t,0,sizeof(t));sz=0; for(int i=1;i<=n;i++) scanf("%s",s+1),ins(s); scanf("%s",s+1); m=strlen(s+1); getAC(); dp(); } }