后缀自动机裸题....
Time Limit: 2000MS | Memory Limit: Unknown | 64bit IO Format: %lld & %llu |
Description
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 fdjskalajfkdsla Output: 3
Notice: new testcases added
Source
#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> using namespace std; const int CHAR=26,maxn=251000; struct SAM_Node { SAM_Node *fa,*next[CHAR]; int len,id,pos; SAM_Node(){} SAM_Node(int _len) { fa=0; len=_len; memset(next,0,sizeof(next)); } }; SAM_Node SAM_node[maxn*2],*SAM_root,*SAM_last; int SAM_size; SAM_Node *newSAM_Node(int len) { SAM_node[SAM_size]=SAM_Node(len); SAM_node[SAM_size].id=SAM_size; return &SAM_node[SAM_size++]; } SAM_Node *newSAM_Node(SAM_Node *p) { SAM_node[SAM_size]=*p; SAM_node[SAM_size].id=SAM_size; return &SAM_node[SAM_size++]; } void SAM_init() { SAM_size=0; SAM_root=SAM_last=newSAM_Node(0); SAM_node[0].pos=0; } void SAM_add(int x,int len) { SAM_Node *p=SAM_last,*np=newSAM_Node(p->len+1); np->pos=len;SAM_last=np; for(;p&&!p->next[x];p=p->fa) p->next[x]=np; if(!p) { np->fa=SAM_root; return ; } SAM_Node *q=p->next[x]; if(q->len==p->len+1) { np->fa=q; return ; } SAM_Node *nq=newSAM_Node(q); nq->len=p->len+1; q->fa=nq; np->fa=nq; for(;p&&p->next[x]==q;p=p->fa) p->next[x]=nq; } void SAM_build(char *s) { SAM_init(); int len=strlen(s); for(int i=0;i<len;i++) SAM_add(s[i]-'a',i+1); } char A[maxn],B[maxn]; int main() { scanf("%s%s",A,B); SAM_build(A); int m=strlen(B),ans=0,temp=0; SAM_Node *now=SAM_root; for(int i=0;i<m;i++) { int c=B[i]-'a'; if(now->next[c]) { now=now->next[c]; temp++; } else { while(now&&!now->next[c]) now=now->fa; if(now) { temp=now->len+1; now=now->next[c]; } else { temp=0; now=SAM_root; } } ans=max(ans,temp); } printf("%d\n",ans); return 0; }
SPOJ 1811LCS Longest Common Substring
时间: 2024-10-07 04:57:07