Longest Common Substring
Time Limit: 8000/4000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5375 Accepted Submission(s): 1910
Problem Description
Given two strings, you have to tell the length of the Longest Common Substring of them.
For example:
str1 = banana
str2 = cianaic
So the Longest Common Substring is "ana", and the length is 3.
For example:
str1 = banana
str2 = cianaic
So the Longest Common Substring is "ana", and the length is 3.
Input
The input contains several test cases. Each test case contains two strings, each string will have at most 100000 characters. All the characters are in lower-case.
Process to the end of file.
Process to the end of file.
Output
For each test case, you have to tell the length of the Longest Common Substring of them.
Sample Input
banana
cianaic
Sample Output
3
这题求最长公共字串,简单的DP并不可以搞定,因为数据范围len<=10^5,所以我用后缀数组做的。
这里将两个字符串连起来,中间用特殊符号隔开,只要两个在排序上连续的后缀的起点在断点两边,就用其更新答案。
1 #include <iostream> 2 #include <cstring> 3 #include <cstdio> 4 5 6 const int maxn=200010; 7 int Ws[maxn],Wv[maxn],Wa[maxn],Wb[maxn],lcp[maxn],sa[maxn],rank[maxn],r[maxn]; 8 bool cmp(int *p,int i,int j,int l) 9 {return p[i]==p[j]&&p[i+l]==p[j+l];} 10 11 void DA(int n,int m) 12 { 13 int i,j,p,*x=Wa,*y=Wb; 14 for(i=0;i<m;i++)Ws[i]=0; 15 for(i=0;i<n;i++)++Ws[x[i]=r[i]]; 16 for(i=1;i<m;i++)Ws[i]+=Ws[i-1]; 17 for(i=n-1;i>=0;i--) 18 sa[--Ws[x[i]]]=i; 19 for(j=1,p=1;p<n;j*=2,m=p) 20 { 21 for(p=0,i=n-j;i<n;i++)y[p++]=i; 22 for(i=0;i<n;i++) 23 if(sa[i]>=j) 24 y[p++]=sa[i]-j; 25 26 for(i=0;i<m;i++)Ws[i]=0; 27 for(i=0;i<n;i++)Wv[i]=x[y[i]]; 28 for(i=0;i<n;i++)++Ws[Wv[i]]; 29 for(i=1;i<m;i++)Ws[i]+=Ws[i-1]; 30 for(i=n-1;i>=0;i--)sa[--Ws[Wv[i]]]=y[i]; 31 for(std::swap(x,y),p=1,i=1,x[sa[0]]=0;i<n;i++) 32 x[sa[i]]=cmp(y,sa[i-1],sa[i],j)?p-1:p++; 33 } 34 } 35 void LCP(int n) 36 { 37 int i,j,k=0; 38 for(i=1;i<=n;i++) 39 rank[sa[i]]=i; 40 41 for(i=0;i<n;lcp[rank[i++]]=k) 42 for(k?--k:k,j=sa[rank[i]-1];r[i+k]==r[j+k];k++); 43 } 44 char str1[maxn>>1],str2[maxn>>1]; 45 int main() 46 { 47 int i,len1,len2,ans; 48 while(~scanf("%s",str1)){ 49 for(len1=0;str1[len1];len1++) 50 r[len1]=str1[len1]; 51 r[len1]='$'; 52 scanf("%s",str2); 53 for(len2=0;str2[len2];len2++) 54 r[len1+len2+1]=str2[len2]; 55 DA(len1+len2+2,128); 56 LCP(len1+len2+1); 57 ans=0; 58 for(i=1;i<len1+len2+1;i++) 59 if(sa[i]<len1&&sa[i-1]>len1||sa[i-1]<len1&&sa[i]>len1) 60 ans=std::max(ans,lcp[i]); 61 printf("%d ",ans); 62 } 63 return 0; 64 }