SPOJ

You are the King of Byteland. Your agents have just intercepted a batch of encrypted enemy messages concerning the date of the planned attack on your island. You immedietaly send for the Bytelandian Cryptographer, but he is currently busy eating popcorn and claims that he may only decrypt the most important part of the text (since the rest would be a waste of his time). You decide to select the fragment of the text which the enemy has strongly emphasised, evidently regarding it as the most important. So, you are looking for a fragment of text which appears in all the messages disjointly at least twice. Since you are not overfond of the cryptographer, try to make this fragment as long as possible.

Input

The first line of input contains a single positive integer t<=10, the number of test cases. t test cases follow. Each test case begins with integer n (n<=10), the number of messages. The next n lines contain the messages, consisting only of between 2 and 10000 characters 'a'-'z', possibly with some additional trailing white space which should be ignored.

Output

For each test case output the length of longest string which appears disjointly at least twice in all of the messages.

Example

Input:14abbabbadabddkabababacabababaOutput:2

(in the example above, the longest substring which fulfills the requirements is 'ba')

 时限开了9s 我tm30ms刚过。。。

#include <stdio.h>#include <string.h>#include <math.h>#include <algorithm>using namespace std;const int N = 1000000+10;typedef long long LL;int sa[N];int t1[N],t2[N],c[N];int rank1[N],height[N];void build_sa(int s[],int n ,int m){    int i,j,p,*x=t1,*y=t2;    for(int i=0; i<m; i++) c[i]=0;    for(int i=0; i<n; i++) c[x[i]=s[i]]++;    for(int i=0; i<m; i++) c[i]+=c[i-1];    for(int i=n-1; i>=0; i--) sa[--c[x[i]]]=i;    for(j=1; j<=n; j<<=1)    {        p=0;        for(i=n-j; i<n; i++) y[p++]=i;        for(i=0; i<n; i++) if(sa[i]>=j) y[p++]=sa[i]-j;        for(i=0; i<m; i++) c[i]=0;        for(i=0; i<n; i++) c[x[y[i]]]++;        for(i=0; i<m; i++) c[i]+=c[i-1];        for(i=n-1; i>=0; i--) sa[--c[x[y[i]]]]=y[i];        swap(x,y);        p=1;        x[sa[0]]=0;        for(i=1; i<n; i++)            x[sa[i]]=(y[sa[i-1]]==y[sa[i]])&&(y[sa[i-1]+j]==y[sa[i]+j])?p-1:p++;        if(p>=n) break;        m=p;    }}int s[N];void getheight(int s[],int n){    int i, j, k=0;    for(i=0; i<=n; i++) rank1[sa[i]]=i;    for(i=0; i<n; i++)    {        if(k) k--;        j=sa[rank1[i]-1];        while(s[i+k]==s[j+k]) k++;        height[rank1[i]]=k;    }    return ;}char str[N];int id[N], pos[N], vis[20], low[20], up[20];int judge(int mid,int m,int n){    int cnt=0;    memset(vis,0,sizeof(vis));    for(int i=1;i<=n;i++)    {        if(height[i]<mid)        {            memset(vis,0,sizeof(vis));            cnt=0;            continue;        }        if(!vis[id[sa[i-1]]])        {            vis[id[sa[i-1]]]=1;            low[id[sa[i-1]]]=up[id[sa[i-1]]]=sa[i-1];        }        if(!vis[id[sa[i]]])        {            vis[id[sa[i]]]=1;            low[id[sa[i]]]=up[id[sa[i]]]=sa[i];        }        if(vis[id[sa[i-1]]]==1)        {            low[id[sa[i-1]]]=min(low[id[sa[i-1]]],sa[i-1]);            up[id[sa[i-1]]]=max(up[id[sa[i-1]]],sa[i-1]);            if(up[id[sa[i-1]]]-low[id[sa[i-1]]]>=mid)            {                vis[id[sa[i-1]]]=2;                cnt++;            }        }        if(vis[id[sa[i]]]==1)        {            low[id[sa[i]]]=min(low[id[sa[i]]],sa[i]);            up[id[sa[i]]]=max(up[id[sa[i]]],sa[i]);            if(up[id[sa[i]]]-low[id[sa[i]]]>=mid)            {                vis[id[sa[i]]]=2;                cnt++;            }        }        if(cnt==m) return 1;    }    return 0;}int main(){    int t;    scanf("%d", &t);    while(t--)    {        int n=0, m, x=2*'z', maxt=0;        scanf("%d", &m);        for(int i=0;i<m;i++)        {            scanf("%s",str);            int len=strlen(str);            maxt=max(maxt,len);            for(int j=0;str[j];j++)            {                s[n]=str[j];id[n]=i;n++;            }            s[n]=++x;id[n]=-1;n++;        }        s[n]=0,id[n]=0;        build_sa(s,n+1,1000);        getheight(s,n);        int l=1,r=maxt,ans=0;        while(l<=r)        {            int mid=(l+r)/2;            if(judge(mid,m,n)) l=mid+1,ans=mid;            else r=mid-1;        }        printf("%d\n",ans);    }    return 0;}