傳送門
程式碼：
先用特殊字元把所有字串連線在一起。
然後二分答案將 $sa$陣列分組。
討論是否存在一個組滿足組內對於每一個字串都存在兩段不相交字串滿足條件。

#include<bits/stdc++.h>
#define ri register int
using namespace std;
const int N=2e5
 
+5;
int n,m,rk[N],ht[N],sa[N],sa2[N],len[N],st[N][18],Log[N],T_T,tt,mx[N],mn[N],idx[N];
char s[N],t[10005];
inline void Sort(){
	static int cnt[N];
	for(ri i=1;i<=m;++i)cnt[i]=0;
	for(ri i=1;i<=n;++i)++cnt[rk[i]];
	for(ri i=2;i<=m;++i)cnt[i]+=cnt[i-1];
	for(ri i=n;i;--i)sa[cnt[rk[sa2[i]]]--]=sa2[
 
i];
}
inline void getsa(){
	for(ri i=1;i<=n;++i)rk[i]=(int)s[i],sa2[i]=i;
	m=122,Sort();
	for(ri w=1,p=0;m^n;w<<=1,p=0){
		for(ri i=n-w+1;i<=n;++i)sa2[++p]=i;
		for(ri i=1;i<=n;++i)if(sa[i]>w)sa2[++p]=sa[i]-w;
		Sort(),swap(sa2,rk),rk[sa[1]]=p=1;
		for(ri i=2;i<=n;++i)rk[sa[i]]=(sa2[
 
sa[i]]==sa2[sa[i-1]]&&sa2[sa[i]+w]==sa2[sa[i-1]+w])?p:++p;
		m=p;
	}
	for(ri i=1,j,k=0;i<=n;ht[rk[i++]]=k)for(k?--k:k,j=sa[rk[i]-1];s[i+k]==s[j+k];++k);
	for(ri i=1;i<=n;++i)st[i][0]=ht[i];
	for(ri j=1;j<=17;++j)for(ri i=1;i+(1<<j)<=n;++i)st[i][j]=min(st[i][j-1],st[i+(1<<(j-1))][j-1]);
}
inline int rmq(int l,int r){return min(st[l][Log[r-l+1]],st[r-(1<<Log[r-l+1])+1][Log[r-l+1]]);}
inline bool solve(int l,int r,int mid){
	for(ri i=1;i<=tt;++i)mx[i]=-0x3f3f3f3f,mn[i]=0x3f3f3f3f;
	for(ri i=l;i<=r;++i)mx[idx[sa[i]]]=max(mx[idx[sa[i]]],len[idx[sa[i]]]-sa[i]+1),mn[idx[sa[i]]]=min(mn[idx[sa[i]]],len[idx[sa[i]]]-sa[i]+1);
	for(ri i=1;i<=tt;++i)if(mx[i]-mn[i]<mid)return 0;
	return 1;
}
inline bool check(int mid){
	for(ri l=1,r=1;l<=n;l=r+1){
		while(ht[l]<mid&&l<=n)++l;
		if(l>n)break;
		r=l;
		while(ht[r+1]>=mid&&r<n)++r;
		if(solve(l-1,r,mid))return 1;
	}
	return 0;
}
int main(){
	freopen("lx.in","r",stdin);
	Log[0]=-1;
	for(ri i=1;i<=200000;++i)Log[i]=Log[i>>1]+1;
	scanf("%d",&T_T);
	while(T_T--){
		scanf("%d",&tt),n=0;
		for(ri i=1;i<=tt;++i){
			scanf("%s",t+1),s[++n]=(char)i,len[i]=strlen(t+1);
			for(ri j=1;j<=len[i];++j)s[++n]=t[j];
			len[i]+=len[i-1]+1;
		}
		for(ri i=1;i<=tt;++i)for(ri j=len[i-1]+1;j<=len[i];++j)idx[j]=i;
		getsa();
		int l=1,r=len[1],ans=0;
		while(l<=r){
			int mid=l+r>>1;
			if(check(mid))ans=mid,l=mid+1;
			else r=mid-1;
		}
		cout<<ans<<'\n';
	}
	return 0;
}