HDU 1403-Longest Common Substring （后缀数组）

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


题目大意：给两个字符串，求最长的公共子串的长度。
题目分析：扫一遍height数组即可。

代码如下：

# include<iostream>
# include<cstdio>
# include<queue>
# include<cmath>
# include<string>
# include<cstring>
# include<algorithm>
using namespace std;

const int N=100000;

int SA[N*2+5];
int cnt[N*2+5];
int rk[N*2+5];
int tSA[N*2+5];
int height[N*2+5];
int n;
string str;

bool same(int i,int j,int k)
{
	if(tSA[i]!=tSA[j]) return false;
	if(i+k>=n&&j+k<n) return false;
	if(i+k<n&&j+k>=n) return false;
	return tSA[i+k]==tSA[j+k];
}

void buildSA(string s)
{
	int m=27;
	n=s.size();
	for(int i=0;i<m;++i) cnt[i]=0;
	for(int i=0;i<n;++i) ++cnt[rk[i]=s[i]-'a'];
	for(int i=1;i<m;++i) cnt[i]+=cnt[i-1];
	for(int i=n-1;i>=0;--i) SA[--cnt[rk[i]]]=i;
	for(int k=1;k<=n;k<<=1){
		int p=0;
		for(int i=n-k;i<n;++i) tSA[p++]=i;
		for(int i=0;i<n;++i) if(SA[i]>=k) tSA[p++]=SA[i]-k;
		
		for(int i=0;i<m;++i) cnt[i]=0;
		for(int i=0;i<n;++i) ++cnt[rk[tSA[i]]];
		for(int i=1;i<m;++i) cnt[i]+=cnt[i-1];
		for(int i=n-1;i>=0;--i) SA[--cnt[rk[tSA[i]]]]=tSA[i];
		
		swap(rk,tSA);
		p=1;
		rk[SA[0]]=0;
		for(int i=1;i<n;++i)
			rk[SA[i]]=same(SA[i],SA[i-1],k)?p-1:p++;
		if(p>=n) break;
		m=p;
	}
}

void getHeight()
{
	for(int i=0;i<n;++i) rk[SA[i]]=i;
	int k=0;
	for(int i=0;i<n;++i){
		if(rk[i]==0){
			height[rk[i]]=k=0;
		}else{
			if(k) --k;
			int j=SA[rk[i]-1];
			while(i+k<n&&j+k<n&&str[i+k]==str[j+k]) ++k;
			height[rk[i]]=k;
		}
	}
}

string str1,str2;

bool diff(int i,int j,int m)
{
	if(SA[i]==m||SA[j]==m) return false;
	return (SA[i]-m)/abs(SA[i]-m)*(SA[j]-m)/abs(SA[j]-m)<0;
}

int f(int m)
{
	int ans=0;
	for(int i=1;i<n;++i){
		if(diff(i,i-1,m)&&height[i]>ans)
			ans=height[i];
	}
	return ans;
}

int main()
{
	while(cin>>str1>>str2)
	{
		str=str1+(char)('z'+1)+str2;
		buildSA(str);
		getHeight();
		//cout<<"here is good"<<endl;
		printf("%d
",f(str1.size()));
	}
	return 0;
}

　　

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