HDU1403 Longest Common Substring

Longest Common Substring

Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3851    Accepted Submission(s): 1443

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

 

Author

Ignatius.L

 

后缀数组的模板题。。。代码加了注释

#include<cstdio>
#include<cstring>
using namespace std;
const int MAXN=1000010;
int wa[MAXN],wb[MAXN],wv[MAXN],ws[MAXN];
int sa[MAXN],r[MAXN<<1];
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;
    //对长度为1的字符串排序
    for(i=0;i<m;i++)
        ws[i]=0;
    for(i=0;i<n;i++)    //统计各个字符的数量
        ws[x[i]=r[i]]++;
    for(i=1;i<m;i++)    //统计不大于i的字符的数量
        ws[i]+=ws[i-1];
    for(i=n-1;i>=0;i--)
        sa[--ws[x[i]]]=i;   //从字符串尾部开始排名，求后缀，字符串长字典序大
    for(j=1,p=1;p<n;j*=2,m=p)
    {
        for(p=0,i=n-j;i<n;i++)  //y里保存的是第二关键字的排序结果，结合图后j很弱字符串的第二关键都是0
            y[p++]=i;
        for(i=0;i<n;i++)    //第二关键字的排序可以直接根据第一次排序的sa来看
            if(sa[i]>=j)    //sa里存的是排名第i的是谁
            y[p++]=sa[i]-j;
        for(i=0;i<n;i++)    //x保存了rank的值，用y又是运用了第二关键字排序的结果，也是基数排序的要求
            wv[i]=x[y[i]];  //y[i]相当于是进行了第二关键字排序的字符的下标
        //对第一关键字基数排序
        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];       //y[i]才是下标
        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++;//rank值相同位置也要相同
    }
    return;
}
int rank[MAXN],height[MAXN];
void callheight(int *r,int *sa,int n)
{
    int i,j,k=0;
    for (i=1;i<=n;++i)  //rank里记录的是排名
        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);    //那个结论height[rank[i]]>=height[rank[i-1]]-1
    return; //k==0不动，不为0就要减一
}
char s[MAXN<<1];
int main()
{
    int i;
    while(scanf("%s",s)!=EOF)
    {
        int len=strlen(s);
        s[len]='0'; //要将两个串合成一条串，然后求他子串的最长公共前缀就行
        int len1=len;
        scanf("%s",s+1+len);
        len=strlen(s);
        for(i=0;i<len;i++)
            r[i]=s[i];
        r[len]=0;
        da(r,sa,len+1,300);
        callheight(r,sa,len);
        int ans=0;
        for(i=2;i<=len;i++)
        {
            if(ans<height[i])
            {
                if((sa[i]>len1&&sa[i-1]<len1)||(sa[i]<len1&&sa[i-1]>len1))  //sa记录的是排名第几的是谁
                    ans=height[i];
            }
        }
        printf("%d\n",ans);
    }
    return 0;
}