HDU 1403 Longest Common Substrung [后缀数组] [LCP] [LCS]

Longest Common Substring

Time Limit: 4000MS

Memory Limit: 32768KB

64bit IO Format: %I64d & %I64u

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

注意有多组数据！！！那么就是用后缀数组求LCS的裸题咯。。

中间用个特殊字符隔开，因为任意两个后缀的LCP必定可以由rank相邻的两个得来，那么只需要求height最大值并且分别在 特殊字符 两边即可。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<vector>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<string>
#include<iomanip>
#include<ctime>
#include<climits>
#include<cctype>
#include<algorithm>
#ifdef WIN32
#define AUTO "%I64d"
#else
#define AUTO "%lld"
#endif
using namespace std;
#define smax(x,tmp) x=max((x),(tmp))
#define smin(x,tmp) x=min((x),(tmp))
#define maxx(x1,x2,x3) max(max(x1,x2),x3)
#define minn(x1,x2,x3) min(min(x1,x2),x3)
const int INF=0x3f3f3f3f;
const int maxn = 200005;
int seq[maxn],cnt[maxn],sa[maxn],t1[maxn],t2[maxn];
inline bool equal(int temp[],int i,int len)
{
return temp[sa[i-1]]==temp[sa[i]] && temp[sa[i-1]+len]==temp[sa[i]+len];
}
void da(int lens,int lim)
{
int *x=t1,*y=t2;
for(int i=0;i<lim;i++) cnt[i]=0;
for(int i=0;i<lens;i++) cnt[ x[i]=seq[i] ]++;
for(int i=1;i<lim;i++) cnt[i] += cnt[i-1];
for(int i=lens-1;i>=0;i--) sa[--cnt[x[i]]] = i;
for(int k=1;k<=lens;k<<=1)
{
int p=-1;
for(int i=lens-k;i<lens;i++) y[++p]=i;
for(int i=0;i<lens;i++) if(sa[i]>=k) y[++p]=sa[i]-k;
for(int i=0;i<lim;i++) cnt[i]=0;
for(int i=0;i<lens;i++) cnt[x[y[i]]]++;
for(int i=1;i<lim;i++) cnt[i] += cnt[i-1];
for(int i=lens-1;i>=0;i--) sa[--cnt[x[y[i]]]] = y[i];
swap(x,y); p=1;
x[sa[0]]=0;
for(int i=1;i<lens;i++)
x[sa[i]] = equal(y,i,k)? p-1 : p++;
if(p>=lens) break;
lim = p;
}
}
int rank[maxn],height[maxn];
void get_height(int lens)
{
int k=0;
for(int i=1;i<=lens;i++) rank[sa[i]]=i;
for(int i=0;i<lens;i++)
{
if(k) k--;
int j=sa[rank[i]-1];
while(seq[i+k]==seq[j+k]) k++;
height[rank[i]]=k;
}
}
char s[maxn];
int work()
{
if(!~scanf("%s",s)) return -1;
memset(seq,0,sizeof(seq));
int len = strlen(s);
int lim = 0;
for(int i=0;i<len;i++) seq[i]=s[i]-'a'+1;
seq[len]='0'; //special character
scanf("%s",s+len+1);
int lens = strlen(s+len+1) + len + 1;
for(int i=len+1;i<lens;i++) seq[i]=s[i]-'a'+1;
da(lens+1,150);
get_height(lens);
int ans = 0;
for(int i=2;i<=lens;i++)
if((sa[i-1]<len&&sa[i]>len) || (sa[i-1]>len&&sa[i]<len))
smax(ans , height[i]);
return ans;
}
int main()
{
freopen("lcs.in","r",stdin);
freopen("lcs.out","w",stdout);
while(true)
{
int ans = work();
if(!~ans) break;
printf("%d\n",ans);
}
return 0;
}