HDU1403-Longest Common Substring

最新推荐文章于 2019-07-27 00:33:22 发布

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本文介绍了一种求解两字符串最长公共子串的问题，并提供了一个详细的算法实现案例。通过构造特殊字符串并利用后缀数组和LCP数组，该算法能在较短时间内找到最长公共子串的长度。

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Longest Common Substring

Time Limit: 8000/4000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6177 Accepted Submission(s): 2185

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

Author

Ignatius.L

题意：给出两个字符串，求出最多可以匹配多少

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <stack>
#include <queue>
#include <vector>
#include <map>

using namespace std;

#define N 200010

struct Sa
{
    char s[N],s1[N],s2[N];
    int rk[2][N],sa[N],h[N],w[N],now,n,len1,len2;
    int rmq[N][20],lg[N];

    bool GetS()
    {
        if(scanf("%s",s1+1)==EOF) return false;
        scanf("%s",s2+1);
        len1=strlen(s1+1);
        len2=strlen(s2+1);
        for(int i=1;i<=len1;i++) s[i]=s1[i];
        s[len1+1]='$';
        for(int i=len1+2;i<=len2+len1+2;i++) s[i]=s2[i-len1-1];
        n=len1+len2+1;
        return true;
    }

    void getsa(int z,int &m)
    {
        int x=now,y=now^=1;
        for(int i=1;i<=z;i++) rk[y][i]=n-i+1;
        for(int i=1,j=z;i<=n;i++)
            if(sa[i]>z) rk[y][++j]=sa[i]-z;
        for(int i=1;i<=m;i++) w[i]=0;
        for(int i=1;i<=n;i++) w[rk[x][rk[y][i]]]++;
        for(int i=1;i<=m;i++) w[i]+=w[i-1];
        for(int i=n;i>=1;i--) sa[w[rk[x][rk[y][i]]]--]=rk[y][i];
        for(int i=m=1;i<=n;i++)
        {
            int *a=rk[x]+sa[i],*b=rk[x]+sa[i-1];
            rk[y][sa[i]]=*a==*b&&*(a+z)==*(b+z)?m-1:m++;
        }
    }

    void getsa(int m)
    {
        now=rk[1][0]=sa[0]=s[0]=0;
        n=strlen(s+1);
        for(int i=1;i<=m;i++) w[i]=0;
        for(int i=1;i<=n;i++) w[s[i]]++;
        for(int i=1;i<=m;i++) rk[1][i]=rk[1][i-1]+(bool)w[i];
        for(int i=1;i<=m;i++) w[i]+=w[i-1];
        for(int i=1;i<=n;i++) rk[0][i]=rk[1][s[i]];
        for(int i=1;i<=n;i++) sa[w[s[i]]--]=i;
        for(int x=1,y=rk[1][m];x<=n&&y<=n;x<<=1) getsa(x,y);
        for(int i=1,j=0;i<=n;h[rk[now][i++]]=j?j--:j)
        {
            if(rk[now][i]==1) continue;
            int k=n-max(sa[rk[now][i]-1],i);
            while(j<=k&&s[sa[rk[now][i]-1]+j]==s[i+j]) ++j;
        }

    }

    void getrmq()
    {
        h[n+1]=h[1]=lg[1]=0;
        for(int i=2;i<=n;i++)
            rmq[i][0]=h[i],lg[i]=lg[i>>1]+1;
        for(int i=1;(1<<i)<=n;i++)
        {
            for(int j=2;j<=n;j++)
            {
                if(j+(1<<i)>n+1) break;
                rmq[j][i]=min(rmq[j][i-1],rmq[j+(1<<i-1)][i-1]);
            }
        }
    }

    int lcp(int x,int y)
    {
        int l=min(rk[now][x],rk[now][y])+1,r=max(rk[now][x],rk[now][y]);
        return min(rmq[l][lg[r-l+1]],rmq[r-(1<<lg[r-l+1])+1][lg[r-l+1]]);
    }

    int work()
    {
        int ans=-1;
        for(int i=2;i<=n;i++)
        {
            int MAX=max(sa[i-1],sa[i]);
            int MIN=min(sa[i-1],sa[i]);
            if(MAX>len1&&MIN<len1&&ans<h[i])
                ans=h[i];
        }
        return ans;
    }
}sa;

int main()
{
    while(sa.GetS())
    {
        sa.getsa(300);
        printf("%d\n",sa.work());
    }
    return 0;
}