Substrings
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 2 3 ABCD BCDFF BRCD 2 rose orchid Sample Output 2 2 Source |
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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 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);
}
}