POJ 1226 出現或者翻轉後出現在每一個串中的子串

Substrings

Time Limit: 1000MS		Memory Limit: 10000K
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 should be one line per test case containing the length of the largest string found.

Sample Input

23ABCDBCDFFBRCD2roseorchid

Sample Output

22 

題意：給定一些字串，求出現或者翻轉後出現在每一個串中的最長子串。

解題思路：將每一個串以及翻轉後的串都連起來，中間用不相同且未出現過的字元分割開，然後二分長度，根據尾碼數組height值分組判定。

代碼：

/* ***********************************************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;}