POJ 1226 substrings (suffix array)

Substrings Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 13919 Accepted: 4922 Description you given a number of case-sensitive strings of alphabetic characters, find the largest string X, such tha T either X, or its inverse can is found as a substring of any of the given strings. Input the first line of the input contains a single integer t (1 <= t <=), the number of test cases, followed by The input data for each test case. The first line of all test case contains a single integer n (1 <= n <=), 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 is 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 Source Tehran 2002 Preliminary

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The longest common substring that appears in each string after the occurrence or reversal.

Puzzle: Suffix Array

Each string is reversed followed by the original string, and then all strings are connected. Then two points can be judged.

#include <iostream> #include <cstdio> #include <algorithm> #include <cstring> #define N 100003
using namespace Std;
int M,n,p,len;
int sa[n],rank[n],height[n],xx[n],yy[n],*x,*y;
int b[n],a[n],vis[n],pos[n];
Char S[n];
	void Init () {memset (sa,0,sizeof (SA));
	memset (rank,0,sizeof (rank));
	Memset (b,0,sizeof (b));
	memset (vis,0,sizeof (VIS));
	memset (Pos,0,sizeof (POS));
	memset (A,0,sizeof (a));
memset (height,0,sizeof (height)); 
} int cmp (int i,int j,int l) {return y[i]==y[j]&& (i+l>len?-1:y[i+l]) = = (J+l>len?-1:y[j+l]); void Get_sa ()
	{x=xx; y=yy; m=500;
	for (int i=1;i<=len;i++) b[x[i]=a[i]]++;
	for (int i=1;i<=m;i++) b[i]+=b[i-1];
	for (int i=len;i>=1;i--) sa[b[x[i]]--]=i;
		for (int k=1;k<=len;k<<=1) {p=0;
		for (int i=len-k+1;i<=len;i++) y[++p]=i;
		for (int i=1;i<=len;i++) if (sa[i]>k) y[++p]=sa[i]-k;
		for (int i=1;i<=m;i++) b[i]=0;
		for (int i=1;i<=len;i++) b[x[y[i]]]++; for (int i=1;i<=m;i++) b[i]+=b[i-1];
		for (int i=len;i>=1;i--) sa[b[x[y[i]]]--]=y[i]; Swap (x, y); p=2;
		X[sa[1]]=1;
		for (int i=2;i<=len;i++) x[sa[i]]=cmp (sa[i-1],sa[i],k)? p-1:p++;
		if (P>len) break;
	m=p+1;
	} p=0;
	for (int i=1;i<=len;i++) rank[sa[i]]=i;
		for (int i=1;i<=len;i++) {if (rank[i]==1) continue;
		int j=sa[rank[i]-1];
		while (I+p<=len&&j+p<=len&&a[i+p]==a[j+p]) p++;
		Height[rank[i]]=p;
	P=max (p-1,0);
	}} int pd (int x) {int size=0; int last=1;
	memset (vis,0,sizeof (VIS)); 
	if (Pos[sa[1]]) size=1,vis[pos[sa[1]]]=1;
	 	for (int i=2;i<=len;i++) if (height[i]>=x) {vis[pos[sa[i]]]++;
	 if (Vis[pos[sa[i]]]==1&&pos[sa[i]]) size++;
	 	} else {if (size==n) return 1;
	 	for (int j=last;j<i;j++) vis[pos[sa[j]]]=0;
		if (Pos[sa[i]]) size=1,vis[pos[sa[i]]]=1;
		else size=0;
	 last=i;
	} if (size==n) return 1;
return 0;
	} int main () {freopen ("a.in", "R", stdin);
	Freopen ("My.out", "w", stdout);
	int T;
	scanf ("%d", &t); for (iNT t=1;t<=t;t++) {init (); scanf ("%d", &n);
		len=0; int base=200;
		int n1=0;
			for (int i=1;i<=n;i++) {scanf ("%s", s+1);
			N1=strlen (s+1);
			if (i!=1) a[++len]=++base;
			for (int j=1;j<=n1;j++) a[++len]=s[j],pos[len]=i;
			A[++len]=++base;
		for (int j=n1;j>=1;j--) a[++len]=s[j],pos[len]=i;
			} if (n==1) {printf ("%d\n", N1);
		Continue
		} Get_sa (); int l=1; int R=len;
		int ans=0;
			while (l<=r) {int mid= (L+R)/2;
			if (PD (mid)) Ans=max (Ans,mid), l=mid+1;
		else r=mid-1;
	} printf ("%d\n", ans); }
}