后缀数组：HDU1043 Longest Common Substring

最新推荐文章于 2020-07-09 11:18:48 发布

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最新推荐文章于 2020-07-09 11:18:48 发布

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原文链接：http://www.cnblogs.com/TenderRun/p/5198379.html

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.

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.

Output

For each test case, you have to tell the length of the Longest Common Substring of them.

Sample Input

banana cianaic

Sample Output

　　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\n",ans);    
62     }
63     return 0;
64 }

转载于:https://www.cnblogs.com/TenderRun/p/5198379.html