poj 1226 Substrings

Description

You are given a number of case-sensitive strings of alphabetic characters, find the largest string X, such that either X, or its inverse can be found as a substring of any of the given strings.

Input

The first line of the input contains a single integer t (1 <= t <= 10), the number of test cases, followed by the input data for each test case. The first line of each test case contains a single integer n (1 <= n <= 100), the number of given strings, followed by n lines, each representing one string of minimum length 1 and maximum length 100. There is no extra white space before and after a string.

Output

There should be one line per test case containing the length of the largest string found.

Sample Input

2
3
ABCD
BCDFF
BRCD
2
rose
orchid

Sample Output

2
2 


思路： 本题和poj2774很像，2774是只有两个串，而这里是有n个串，这题也可以暴力kmp去做。当然了，用后缀数组来做这题更好，由于h数组的特殊性质，这题我们可以把每个字符串正向
和反向连在一起，每个字符串中间都间隔一个特殊的不同的字符。 然后求出sa数组和h数组以后，我们就二分枚举答案，从h[1],h[2]不断枚举下去，看是否比mid大，如果大于等于mid，
我们就看他们所在的原位置是否属于原来的那n个字符串，如果筹齐了n个字符串，则说明mid可行。  

#include<cstdio>
#include<cmath>
#include<iostream>  
#include<algorithm>  
#include<cstring>  
using namespace std;  
int const N=20000+1000;  
int n,t[N],num[N],wa[N<<1],wb[N<<1],wv[N],sa[N],rk[N],h[N];  
int  s[N];   
char s1[103];  
int cmp(int *r,int x,int y,int z){
    return r[x]==r[y] && r[x+z]==r[y+z];  
}
void build_sa(int   *r,int *sa,int n,int m){
    int i,j,p,*x=wa,*y=wb;  
    for(i=0;i<m;i++) num[i]=0; 
    for(i=0;i<n;i++) num[x[i]=r[i]]++;  
    for(i=1;i<m;i++) num[i]+=num[i-1];  
    for(i=n-1;i>=0;i--) sa[--num[x[i]]]=i;    
    for(p=1,j=1;p<n;j<<=1,m=p){
        for(p=0,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++) num[i]=0;  
        for(i=0;i<n;i++) wv[i]=x[y[i]];  
        for(i=0;i<n;i++) num[wv[i]]++; 
        for(i=1;i<m;i++) num[i]+=num[i-1];  
        for(i=n-1;i>=0;i--) sa[--num[wv[i]]]=y[i];  
        swap(x,y);  
        for(p=1,x[sa[0]]=0,i=1;i<n;i++) 
            x[sa[i]]=cmp(y,sa[i-1],sa[i],j)? p-1:p++;  
    }
    for(i=0;i<n;i++) rk[i]=x[i];  
}
void build_h(int   *r,int *sa,int n){
    int k=0;  
    for(int i=0;i<n;i++){
        if(k) k--;  
        int j=sa[rk[i]-1];  
        while ( r[i+k]==r[j+k])  k++; 
        h[rk[i]]=k;  
    }
} 
int check(int mid,int m){
    int vis[101],num=0;  
    memset(vis,0,sizeof(vis));  
    for(int i=1;i<=m;i++){
        if(h[i]>=mid){
            if(!vis[t[sa[i]]]){
                num++;vis[t[sa[i]]]=1;  
            }
            if(!vis[t[sa[i-1]]]){
                num++; vis[t[sa[i-1]]]=1;  
            }
            if(num==n) return 1;  
        }else {
            num=0;             
            memset(vis,0,sizeof(vis));  
        }
    }
    return 0; 
}
int main(){
    int cas;  
    scanf("%d",&cas); 
    while (cas--){
        scanf("%d",&n);  
        int m=0,tot=130;  
        for(int i=0;i<n;i++){
            scanf("%s",s1);  
            int len=strlen(s1);  
            for(int j=0;j<len;j++) s[m++]=s1[j],t[m-1]=i+1;  
            s[m++]=tot++;  
            for(int j=len-1;j>=0;j--) s[m++]=s1[j],t[m-1]=i+1; 
            s[m++]=tot++;  
        }
        s[m]=0;  
        build_sa(s,sa,m+1,330);   
        build_h(s,sa,m);  
        int l=0,r=100,ans=0;  
        while (l<=r){
            int mid=(l+r)/2;  
            if(check(mid,m)){
                ans=mid; l=mid+1;  
            }else r=mid-1;  
        }
        printf("%d\n",ans); 
    }
    return 0;
}