poj 1226 Substrings （后缀数组）

Substrings

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 13919 Accepted: 4922

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 

Source

Tehran 2002 Preliminary

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题目大意：出现或反转后出现在每个字符串中的最长公共子串。

题解：后缀数组

将每个串反转后接在原串后面，再将所有串相连。然后二分判定即可。

#include<iostream>#include<cstdio>#include<algorithm>#include<cstring>#define N 100003using 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);}}