DNA repair

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 2612    Accepted Submission(s): 1406 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.   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.   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 2008 Asia Hefei Regional Contest Online by USTC

给你n个模式串，再给一个字符串s，问最少改变s的几个字符，使s不包含这n个模式串。

首先，根据所给字符串构造trie图。

之后，可以DP了。

dp[i][j]表示字符串长度为i，状态为自动机上状态j时的最少改动字符数。

则dp[i][j]=min(dp[i-1][k]),其中j状态可在自动机上由k状态转化而来。

边界条件：dp[0][0]=0, 0为自动机上根的状态。

#include <cstdio> #include <iostream> #include <string.h> #include <string> #include <queue> #define mem0(a) memset(a,0,sizeof(a)) #define meminf(a) memset(a,0x3f,sizeof(a)) using namespace std; typedef long long ll; const int maxn=1005,inf=0x3f3f3f3f,maxk=4; char s[maxn]; char d[4]; int dp[maxn][maxn]; int id['Z'+1]; int num; struct node{ struct node *fail; struct node *next[maxk]; int cnt,id; void init() { fail=NULL; for (int i=0;i<maxk;i++) next[i]=NULL; cnt=0; id=++num; } }; node *a[maxn],*q[maxn]; void insert(string s,node *root,int len) { node *now=root; int i; for (i=0;i<len;i++) { int pos=id[s[i]]; if (now->next[pos]==NULL) { now->next[pos]=new node; now->next[pos]->init(); a[num]=now->next[pos]; } now=now->next[pos]; } now->cnt++; } void buildfail(node *root) { node *p=root; int front=0,tail=0,i; for (i=0;i<maxk;i++) { if (p->next[i]!=NULL) { p->next[i]->fail=root; q[tail++]=p->next[i]; } else p->next[i]=root; } while (front<tail) { p=q[front]; for (i=0;i<maxk;i++) { if (p->next[i]==NULL) p->next[i]=p->fail->next[i]; else { node *x=p->fail; while (x!=NULL) { if (x->next[i]!=NULL) { p->next[i]->fail=x->next[i]; break; } x=x->fail; } if (p->fail->next[i]->cnt) p->next[i]->cnt++; q[tail++]=p->next[i]; } } front++; } } int main() { int cas=0,n; scanf("%d",&n); id['A']=0;id['T']=1;id['C']=2;id['G']=3; d[0]='A';d[1]='T';d[2]='C';d[3]='G'; while (n) { int i,len,j,k; cas++; num=0; node *root=new node; root->init(); a[1]=root; for (i=1;i<=n;i++) { scanf("%s",s); insert(s,root,strlen(s)); } buildfail(root); scanf("%s",s); len=strlen(s); meminf(dp); dp[0][1]=0; for (i=1;i<=len;i++) { for (j=1;j<=num;j++) { if (dp[i-1][j]==inf) continue; for (k=0;k<4;k++) { if (a[j]->next[k]->cnt==0) { dp[i][a[j]->next[k]->id]= min(dp[i][a[j]->next[k]->id], dp[i-1][j]+(d[k]==s[i-1]?0:1)); } } } } int ans=inf; for (i=1;i<=num;i++) ans=min(ans,dp[len][i]); if (ans==inf) ans=-1; printf("Case %d: %d\n",cas,ans); scanf("%d",&n); } return 0; }