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

输入样例#1： 
alsdfkjfjkdsal fdjskalajfkdsla aaaajfaaaa 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; }

　　

 

转载于:https://www.cnblogs.com/JYYHH/p/8458206.html

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