After POJ 1226 appears or is flipped, the substring appears in each string.

Source: Internet
Author: User

Substrings

Time Limit:1000 MS   Memory Limit:10000 K
Total Submissions:10998   Accepted:3824

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

Sample Input

23ABCDBCDFFBRCD2roseorchid

Sample Output

22 


Given some strings, find the largest substring in each string after it appears or is flipped.

Solution: connect each string and the flipped string, separate them with different and unused characters in the middle, and divide them into two parts based on the height value of the suffix array.

Code:

/* ***********************************************Author :xianxingwuguanCreated Time :2014-2-1 2:59:27File Name :1.cpp************************************************ */#pragma comment(linker, "/STACK:102400000,102400000")#include <stdio.h>#include <string.h>#include <iostream>#include <algorithm>#include <vector>#include <queue>#include <set>#include <map>#include <string>#include <math.h>#include <stdlib.h>#include <time.h>using namespace std;const int maxn=30100;const double pi =acos(-1.0);const double eps =1e-8;int height[maxn],sa[maxn],rank[maxn],c[maxn],t1[maxn],t2[maxn];    void da(int *str,int n,int m)    {          int i,j,k,p,*x=t1,*y=t2;          for(i=0;i<m;i++)c[i]=0;          for(i=0;i<n;i++)c[x[i]=str[i]]++;          for(i=1;i<m;i++)c[i]+=c[i-1];          for(i=n-1;i>=0;i--)sa[--c[x[i]]]=i;          for(k=1;k<=n;k<<=1)          {             p=0;             for(i=n-k;i<n;i++)y[p++]=i;             for(i=0;i<n;i++)if(sa[i]>=k)y[p++]=sa[i]-k;             for(i=0;i<m;i++)c[i]=0;             for(i=0;i<n;i++)c[x[y[i]]]++;             for(i=1;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]]==y[sa[i-1]]&&y[sa[i]+k]==y[sa[i-1]+k]?p-1:p++;             if(p>=n)break;             m=p;          }    }    void calheight(int *str,int n)    {          int i,j,k=0;          for(i=0;i<=n;i++)rank[sa[i]]=i;          for(i=0;i<n;i++)          {             if(k)k--;             j=sa[rank[i]-1];             while(str[i+k]==str[j+k])k++;             height[rank[i]]=k;          }             // printf("sa:");for(i=0;i<=n;i++)printf("%d ",sa[i]);puts("");           //   printf("rank:");for(i=0;i<=n;i++)printf("%d ",rank[i]);puts("");             // printf("height:");for(i=0;i<=n;i++)printf("%d ",height[i]);puts("");            }    int str[maxn],loc[maxn],vis[maxn];char ss[maxn];int n;bool check(int mid,int len){      int i,j,tot;      tot=0;      memset(vis,0,sizeof(vis));      for(i=2;i<=len;i++)    {          if(height[i]<mid) {              memset(vis,0,sizeof(vis));              tot=0;          }          else {              if(!vis[loc[sa[i-1]]])     {                  vis[loc[sa[i-1]]]=1;                  tot++;              }              if(!vis[loc[sa[i]]]){                  vis[loc[sa[i]]]=1;                  tot++;              }              if(tot==n)return 1;        }      }      return 0;  }  int main(){      //freopen("data.in","r",stdin);      //freopen("data.out","w",stdout);      int T,i,j;      scanf("%d",&T);      while(T--)      {     scanf("%d",&n);     int len=0;     int ff=200;     int left=0,right;     for(i=1;i<=n;i++)     {    scanf("%s",ss);    int cnt=strlen(ss);    right=cnt;    for(j=0;j<cnt;j++)str[len]=ss[j]+500,loc[len++]=i;    str[len]=ff;    loc[len++]=ff++;    for(j=cnt-1;j>=0;j--)str[len]=ss[j]+500,loc[len++]=i;                  str[len]=ff;    loc[len++]=ff++;     }     str[len]=0;    // cout<<"len="<<len<<endl;     da(str,len+1,1000);     calheight(str,len);     while(left<right)     {    int mid=(left+right+1)>>1;                  if(check(mid,len))left=mid;    else right=mid-1;     }     printf("%d\n",left);      }      return 0;}


Contact Us

The content source of this page is from Internet, which doesn't represent Alibaba Cloud's opinion; products and services mentioned on that page don't have any relationship with Alibaba Cloud. If the content of the page makes you feel confusing, please write us an email, we will handle the problem within 5 days after receiving your email.

If you find any instances of plagiarism from the community, please send an email to: info-contact@alibabacloud.com and provide relevant evidence. A staff member will contact you within 5 working days.

A Free Trial That Lets You Build Big!

Start building with 50+ products and up to 12 months usage for Elastic Compute Service

  • Sales Support

    1 on 1 presale consultation

  • After-Sales Support

    24/7 Technical Support 6 Free Tickets per Quarter Faster Response

  • Alibaba Cloud offers highly flexible support services tailored to meet your exact needs.