hdu 1403 Longest Common Substring 后缀数组求最长公共子串

Longest Common Substring

Time Limit: 8000/4000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 5823 Accepted Submission(s): 2063

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

3

题意：给两个字符串，求最长公共子串。

这一题数据量很大，如果使用dp的方法求复杂度是o(n^2)，显然会超时。这个时候我们使用后缀数组，由于所有子串分别为母串某一个后缀的前缀，所以只需将两串组合在一起形成一条母串，然后求最大的height[i]即可。入门后缀数组应用。

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn=200050;
char s1[maxn],s2[maxn],str[maxn];
int sa[maxn],r[maxn];
int wa[maxn],wb[maxn],wv[maxn];
int ws[maxn];
int cmp(int *r,int a,int b,int l)
{return r[a]==r[b]&&r[a+l]==r[b+l];}
void da(int *r,int *sa,int n,int m)
{
int i,j,p,*x=wa,*y=wb,*t;
for(i=0;i<m;i++) ws[i]=0;
for(i=0;i<n;i++) ws[x[i]=r[i]]++;
for(i=1;i<m;i++) ws[i]+=ws[i-1];
for(i=n-1;i>=0;i--) sa[--ws[x[i]]]=i;
for(j=1,p=1;p<n;j*=2,m=p)
{
for(p=0,i=n-j;i<n;i++) y[p++]=i;
for(i=0;i<n;i++) if(sa[i]>=j) y[p++]=sa[i]-j;
for(i=0;i<n;i++) wv[i]=x[y[i]];
for(i=0;i<m;i++) ws[i]=0;
for(i=0;i<n;i++) ws[wv[i]]++;
for(i=1;i<m;i++) ws[i]+=ws[i-1];
for(i=n-1;i>=0;i--) sa[--ws[wv[i]]]=y[i];
for(t=x,x=y,y=t,p=1,x[sa[0]]=0,i=1;i<n;i++)
x[sa[i]]=cmp(y,sa[i-1],sa[i],j)?p-1:p++;
}
return;
}
int Rank[maxn],height[maxn];
void calheight(int *r,int *sa,int n)
{
int i,j,k=0;
for(i=1;i<=n;i++) Rank[sa[i]]=i;
for(i=0;i<n;height[Rank[i++]]=k)
for(k?k--:0,j=sa[Rank[i]-1];r[i+k]==r[j+k];k++);
return;
}
int ans;

int main()
{
while(scanf("%s",str)!=EOF)
{
/*int len1=strlen(s1);
scanf("%s",str);
//int len2=strlen(s2);
//strcat(str,s1);
str[len1]='0';
scanf("%s",str+len1+1);
//strcat(str,s2);
//str[len1+len2+1]='\0';
int n=strlen(str);
//puts(str);*/
int n=strlen(str);
int len1=n;
str
='9';
scanf("%s",str+len1+1);
//puts(str);
n=strlen(str);
for(int i=0;i<n;i++)
r[i]=str[i];
r
=0;
da(r,sa,n,300);
calheight(r,sa,n);
/*for(int i=0;i<=n;i++)
{
printf("%d %d\n",sa[i],height[i]);
}*/
int Max=-1;
for(int i=2;i<n;i++)
{
if(((sa[i]>=0&&sa[i]<len1)&&(sa[i-1]>len1&&sa[i-1]<n))||((sa[i-1]>=0&&sa[i-1]<len1)&&(sa[i]>len1&&sa[i]<n)))
{
Max=max(Max,height[i]);
}
}
printf("%d\n",Max);
}
}