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SPOJ Problem Set (classical)1812. Longest Common Substring IIProblem code: LCS2 |
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 a bit harder, for some given strings, find the length of the longest common substring of them.
Here common substring means a substring of two or more strings.
The input contains at most 10 lines, each line consists of no more than 100000 lowercase letters, representing a string.
The length of the longest common substring. If such string doesn‘t exist, print "0" instead.
Input: alsdfkjfjkdsal fdjskalajfkdsla aaaajfaaaa Output: 2
Notice: new testcases added
Added by: | Bin Jin |
Date: | 2007-09-24 |
Time limit: | 2s |
Source limit: | 50000B |
Memory limit: | 256MB |
Cluster: | Pyramid (Intel Pentium III 733 MHz) |
Languages: | All except: C++ 4.0.0-8 |
第一个串建SAM,剩下的串在上面跑....记录下每个节点的最小匹配长度(初始值是 len 即能表示的最长后缀),
从后向前更新fa节点的LCS....
#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> using namespace std; const int maxn=110000; struct SAM_Node { SAM_Node *fa,*next[26]; 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; } char str[maxn],other[maxn]; int LCS[maxn*2],len,n; int c[maxn*2]; SAM_Node *top[maxn*2]; int main() { scanf("%s",str); len=strlen(str); SAM_init(); for(int i=0;i<len;i++) SAM_add(str[i]-'a',i+1); for(int i=0;i<SAM_size;i++) c[SAM_node[i].len]++; for(int i=1;i<=len;i++) c[i]+=c[i-1]; for(int i=0;i<SAM_size;i++) top[--c[SAM_node[i].len]]=&SAM_node[i]; while(scanf("%s",other)!=EOF) { n++; int len2=strlen(other); SAM_Node* now=SAM_root; int temp=0; for(int i=0;i<len2;i++) { int x=other[i]-'a'; if(now->next[x]) { temp++; now=now->next[x]; } else { while(now&&now->next[x]==0) now=now->fa; if(now) { temp=now->len+1; now=now->next[x]; } else { temp=0; now=SAM_root; } } if(temp>LCS[now->id]) LCS[now->id]=temp; } for(int i=SAM_size-1;i>=0;i--) { if(LCS[top[i]->id]<top[i]->len) top[i]->len=LCS[top[i]->id]; if(top[i]->fa&&LCS[top[i]->fa->id]<LCS[top[i]->id]) LCS[top[i]->fa->id]=LCS[top[i]->id]; LCS[top[i]->id]=0; } } int ans=0; for(int i=1;i<SAM_size;i++) { if(ans<SAM_node[i].len) ans=SAM_node[i].len; } printf("%d\n",ans); return 0; }
SPOJ LCS2 1812. Longest Common Substring II,布布扣,bubuko.com
SPOJ LCS2 1812. Longest Common Substring II
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原文地址:http://blog.csdn.net/ck_boss/article/details/37124159