Substrings
Time Limit:1000 MS |
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Memory Limit:10000 K |
Total Submissions:10998 |
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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;}