Spoj LCS2 - Longest Common Substring II
题目描述
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.
输入输出格式
输入格式:
The input contains at most 10 lines, each line consists of no more than 100000 lowercase letters, representing a string.
输出格式:
The length of the longest common substring. If such string doesn't exist, print "0" instead.
输入输出样例
输出样例#1:
2
大概就是要你求最多10个串的最长公共子串,裸上后缀自动机。。。。
#include<iostream> #include<cmath> #include<cstdio> #include<cstdlib> #include<cstring> #include<algorithm> #define ll long long #define maxn 4000005 using namespace std; int a[maxn],c[maxn],tot; int base=1,cnt=1,pre=1,n; int f[maxn],ch[maxn][26]; int l[maxn],siz[maxn],ans=0; char s[maxn]; inline void ins(int x){ int p=pre,np=++cnt; pre=np,l[np]=l[p]+1; siz[np]=base; for(;p&&!ch[p][x];p=f[p]) ch[p][x]=np; if(!p) f[np]=1; else{ int q=ch[p][x]; if(l[q]==l[p]+1) f[np]=q; else{ int nq=++cnt; l[nq]=l[p]+1; memcpy(ch[nq],ch[q],sizeof(ch[q])); f[nq]=f[q]; f[q]=f[np]=nq; for(;ch[p][x]==q;p=f[p]) ch[p][x]=nq; } } } inline void build(){ n=strlen(s),tot+=n; for(int i=0;i<n;i++) ins(s[i]-'a'); } inline void solve(){ base--; for(int i=1;i<=cnt;i++) c[l[i]]++; for(int i=tot;i>=0;i--) c[i]+=c[i+1]; for(int i=1;i<=cnt;i++) a[c[l[i]]--]=i; for(int i=1;i<=cnt;i++){ int now=a[i]; siz[f[now]]|=siz[now]; if(siz[now]==base) ans=max(ans,l[now]); } } int main(){ while(scanf("%s",s)==1) build(),base<<=1; solve(); printf("%d\n",ans); return 0; }
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