SPOJ - LCS 后缀自动机入门

LCS - Longest Common Substring

A string is finite sequence of characters over a non-empty finite set Σ.

In this problem, Σ is the set of lowercase letters.

Substring, also called factor, is a consecutive sequence of characters occurrences at least once in a string.

Now your task is simple, for two given strings, find the length of the longest common substring of them.

Here common substring means a substring of two or more strings.

Input

The input contains exactly two lines, each line consists of no more than 250000 lowercase letters, representing a string.

Output

The length of the longest common substring. If such string doesn't exist, print "0" instead.

Example

Input:
alsdfkjfjkdsal
fdjskalajfkdslaOutput:
3

Notice: new testcases added

题意：

　　求两个串的最长公共子串。

题解：

　　先用第一个串构造出后缀自动机，然后逐个的匹配第二个串

　　如果当前节点失配，利用Suffix Links 往后跳就可以了。

#include <bits/stdc++.h>
inline long long read(){long long x=0,f=1;char ch=getchar();while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}return x*f;}
using namespace std;const int N = 3e5+7;const long long mod = 1000000007;int isPlus[N * 2],endpos[N * 2];int d[N * 2];
int tot,slink[2*N],trans[2*N][28],minlen[2*N],maxlen[2*N],pre;
int newstate(int _maxlen,int _minlen,int* _trans,int _slink){maxlen[++tot]=_maxlen;minlen[tot]=_minlen;slink[tot]=_slink;if(_trans)for(int i=0;i<26;i++)trans[tot][i]=_trans[i],d[_trans[i]]+=1;return tot;
}
int add_char(char ch,int u){int c=ch-'a',v=u;int z=newstate(maxlen[u]+1,-1,NULL,0);isPlus[z] = 1;while(v&&!trans[v][c]){trans[v][c]=z;d[z]+=1;v=slink[v];}if(!v){ minlen[z]=1;slink[z]=1;return z;}int x=trans[v][c];if(maxlen[v]+1==maxlen[x]){slink[z]=x;minlen[z]=maxlen[x]+1;return z;}int y=newstate(maxlen[v]+1,-1,trans[x],slink[x]);slink[z]=slink[x]=y;minlen[x]=minlen[z]=maxlen[y]+1;while(v&&trans[v][c]==x){trans[v][c]=y;d[x]--,d[y]++;v=slink[v];}minlen[y]=maxlen[slink[y]]+1;return z;
}
void init_sam() {for(int i = 1; i <= tot; ++i)for(int j = 0; j < 26; ++j) trans[i][j] = 0;pre = tot = 1;
}
char a[N],b[N];
int main() {scanf("%s%s",a,b);init_sam();int n = strlen(a);int m = strlen(b);for(int i = 0; i < n; ++i)pre = add_char(a[i],pre);int ans = 0;int now = 1;int sum = 0;for(int i = 0; i < m; ++i) {if(trans[now][b[i]-'a']) {now = trans[now][b[i]-'a'];} else {while(now) {now = slink[now];if(trans[now][b[i]-'a']) break;}if(!now) now = 1,sum = 0;else {if(trans[now][b[i]-'a']) {sum = maxlen[now] + 1;now = trans[now][b[i]-'a'];}}}ans = max(ans,sum);}printf("%d\n",ans);
}

转载于:https://www.cnblogs.com/zxhl/p/7616318.html