不小于k个字符串中的最长子串
Description
You may have wondered why most extraterrestrial life forms resemble humans, differing by superficial traits such as height, colour, wrinkles, ears, eyebrows and the like. A few bear no human resemblance; these typically have geometric or amorphous shapes like cubes, oil slicks or clouds of dust.
The answer is given in the 146th episode of Star Trek - The Next Generation, titled The Chase. It turns out that in the vast majority of the quadrant's life forms ended up with a large fragment of common DNA.
Given the DNA sequences of several life forms represented as strings of letters, you are to find the longest substring that is shared by more than half of them.
Input
Standard input contains several test cases. Each test case begins with 1 ≤ n ≤ 100, the number of life forms. n lines follow; each contains a string of lower case letters representing the DNA sequence of a life form. Each DNA sequence contains at least one and not more than 1000 letters. A line containing 0 follows the last test case.
Output
For each test case, output the longest string or strings shared by more than half of the life forms. If there are many, output all of them in alphabetical order. If there is no solution with at least one letter, output "?". Leave an empty line between test cases.
Sample Input
3 abcdefg bcdefgh cdefghi 3 xxx yyy zzz 0
Sample Output
bcdefg cdefgh ?
Source
Sol
考虑广义SAM
维护出每一个点出现次数。
暴力把所有可能的点拿出来，二分hash sort就好了
#include<cstdio> #include<iostream> #include<cstdlib> #include<cstring> #include<algorithm> #include<cmath> #define maxn 2000006 #define ll unsigned long long #define P 793999 using namespace std; int n,rt,la,cnt,len[105]; char ch[105][5005]; int num[maxn],now[maxn]; struct sam{ int Max,par,nex[26]; int id,x; }s[maxn]; struct node{ int len; char ch[5005];ll h[5005]; void H(){ for(int j=1;j<=len;j++)h[j]=h[j-1]*P+ch[j]; } }ans[10005]; bool operator ==(node A,node B){ if(A.len!=B.len)return 0; return A.h[A.len]==B.h[B.len]; } int find(node A,node B){ int M=min(A.len,B.len); if(A.ch[1]!=B.ch[1])return 0; int l=1,r=M; while(l<r){ int mid=(l+r+1)>>1; if(A.h[mid]==B.h[mid])l=mid; else r=mid-1; } return l; } bool operator <(node A,node B){ int x=find(A,B); if(min(A.len,B.len)==x)return A.len<B.len; return A.ch[x+1]<B.ch[x+1]; } void init(){ for(int i=1;i<=cnt;i++){ for(int j=0;j<26;j++)s[i].nex[j]=0; s[i].Max=s[i].par=0; num[i]=now[i]=0; s[i].id=s[i].x=0; } rt=la=cnt=1; } void ins(int c,int x){ int np=++cnt,p=la;la=np;s[np].Max=s[p].Max+1; s[np].id=s[np].Max;s[np].x=x; for(;p&&!s[p].nex[c];p=s[p].par)s[p].nex[c]=np; if(!p)s[np].par=rt; else { int q=s[p].nex[c],nq; if(s[q].Max==s[p].Max+1)s[np].par=q; else { nq=++cnt;s[nq].Max=s[p].Max+1; s[nq].id=s[np].id;s[nq].x=x; for(int j=0;j<26;j++)s[nq].nex[j]=s[q].nex[j]; s[nq].par=s[q].par;s[q].par=s[np].par=nq; for(;p&&s[p].nex[c]==q;p=s[p].par)s[p].nex[c]=nq; } } } int main(){ while(~scanf("%d",&n)){ if(!n)break; init(); for(int i=1;i<=n;i++){ scanf(" %s",ch[i]+1); len[i]=strlen(ch[i]+1); la=rt; for(int j=1;j<=len[i];j++)ins(ch[i][j]-'a',i); cout<<len[i]<<endl; } for(int i=1;i<=n;i++){ int k=rt; for(int j=1;j<=len[i];j++){ k=s[k].nex[ch[i][j]-'a']; int p=k; for(;now[p]!=i&&p;p=s[p].par)now[p]=i,num[p]++; } } int Max=0; for(int i=1;i<=cnt;i++){ if(num[i]*2>n)Max=max(Max,s[i].Max); } int tp=0; for(int i=2;i<=cnt;i++){ if(s[i].Max==Max&&num[i]*2>n){ int l=0;tp++; for(int j=s[i].id-Max+1;j<=s[i].id;j++){ ans[tp].ch[++l]=ch[s[i].x][j]; } ans[tp].len=l;ans[tp].H(); } } if(!tp)puts("?"); else { sort(ans+1,ans+tp+1); for(int i=1;i<=tp;i++){ if(ans[i]==ans[i-1])continue; for(int j=1;j<=ans[i].len;j++)printf("%c",ans[i].ch[j]);puts(""); } } puts(""); } return 0; }