题解：
构建出trie图，令f[i][j]表示到第i个字符走到j号节点最少需要修改的字符数，然后枚举后继节点转移即可。
代码：没写caseWA了n发。。。
1 #include<cstdio> 2 #include<cstdlib> 3 #include<cmath> 4 #include<cstring> 5 #include<algorithm> 6 #include<iostream> 7 #include<vector> 8 #include<map> 9 #include<set> 10 #include<queue> 11 #include<string> 12 #define inf 1000000000 13 #define maxn 2000+5 14 #define maxm 500+100 15 #define eps 1e-10 16 #define ll long long 17 #define pa pair<int,int> 18 #define for0(i,n) for(int i=0;i<=(n);i++) 19 #define for1(i,n) for(int i=1;i<=(n);i++) 20 #define for2(i,x,y) for(int i=(x);i<=(y);i++) 21 #define for3(i,x,y) for(int i=(x);i>=(y);i--) 22 #define for4(i,x) for(int i=head[x],y;i;i=e[i].next) 23 #define mod 1000000007 24 using namespace std; 25 inline int read() 26 { 27 int x=0,f=1;char ch=getchar(); 28 while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} 29 while(ch>='0'&&ch<='9'){x=10*x+ch-'0';ch=getchar();} 30 return x*f; 31 } 32 int n,tot,f[maxn][maxn],t[maxn][5],go[maxn]; 33 char s[maxn]; 34 bool v[maxn]; 35 queue<int>q; 36 inline int id(char ch) 37 { 38 if(ch=='A')return 0; 39 if(ch=='G')return 1; 40 if(ch=='C')return 2; 41 if(ch=='T')return 3; 42 } 43 inline void insert() 44 { 45 memset(s,0,sizeof(s)); 46 scanf("%s",s+1);int len=strlen(s+1),now=1; 47 for1(i,len) 48 { 49 int x=id(s[i]); 50 if(!t[now][x])t[now][x]=++tot; 51 now=t[now][x]; 52 } 53 v[now]=1; 54 } 55 void bfs() 56 { 57 q.push(1); 58 while(!q.empty()) 59 { 60 int x=q.front(),y,j;v[x]|=v[go[x]];q.pop(); 61 for0(i,3) 62 { 63 j=go[x]; 64 while(j&&!t[j][i])j=go[j]; 65 if(t[x][i]) 66 { 67 go[y=t[x][i]]=j?t[j][i]:1; 68 q.push(y); 69 }else t[x][i]=j?t[j][i]:1; 70 } 71 } 72 } 73 int main() 74 { 75 freopen("input.txt","r",stdin); 76 freopen("output.txt","w",stdout); 77 int cs=0; 78 while(cin>>n&&n) 79 { 80 tot=1; 81 memset(t,0,sizeof(t)); 82 memset(v,0,sizeof(v)); 83 memset(go,0,sizeof(go)); 84 for1(i,n)insert(); 85 bfs(); 86 memset(s,0,sizeof(s)); 87 scanf("%s",s+1);n=strlen(s+1); 88 for0(i,n)for1(j,tot)f[i][j]=inf; 89 f[0][1]=0; 90 for0(i,n-1) 91 for1(j,tot) 92 if(f[i][j]!=inf) 93 for0(k,3) 94 if(!v[t[j][k]]) 95 f[i+1][t[j][k]]=min(f[i+1][t[j][k]],f[i][j]+(k!=id(s[i+1]))); 96 int ans=inf; 97 for1(i,tot)ans=min(ans,f[n][i]); 98 printf("Case %d: %d\n",++cs,ans==inf?-1:ans); 99 } 100 return 0; 101 }