poj 1226 Substrings （后缀数组）

最新推荐文章于 2020-01-27 10:07:07 发布

clover_hxy

最新推荐文章于 2020-01-27 10:07:07 发布

阅读量525

点赞数

CC 4.0 BY-SA版权

分类专栏：后缀数组

本文链接：https://blog.youkuaiyun.com/clover_hxy/article/details/53931586

后缀数组专栏收录该内容

27 篇文章

订阅专栏

本文介绍了一种利用后缀数组解决寻找多个字符串中出现或反转后出现的最长公共子串的问题。通过将字符串进行特殊处理并连接后，运用二分查找及后缀数组技术高效地找出最长公共子串。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Substrings

Time Limit: 1000MS		Memory Limit: 10000K
Total 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

2
3
ABCD
BCDFF
BRCD
2
rose
orchid

Sample Output

2
2

Source

Tehran 2002 Preliminary

[Submit] [Go Back] [Status] [Discuss]

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);
	}
}