Hdu1403 Longest Common Substring

Longest Common Substring

Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7335    Accepted Submission(s): 2580

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

题目大意：求两个字符串的后缀的LCP.

分析：后缀数组的经典应用.

　　　把两个字符串拼起来，中间用一个特殊字符隔开.对这一整体字符串求height数组，如果height[i]比ans大，并且sa[i-1],sa[i]分布在两个字符串中，就更新ans.

　　　注意多组数据数组清空的问题！我没有清空fir和sec数组，导致一直RE.

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

const int maxn = 1000010;
char s1[maxn],s2[maxn],s[maxn];
int len1,len2,len,sett[maxn],a[maxn],cnt,fir[maxn],sec[maxn],tong[maxn],pos[maxn];
int rk[maxn],sa[maxn],ht[maxn],ans,poss;

void solve()
{
    memset(rk,0,sizeof(rk));
    memset(sa,0,sizeof(sa));
    memset(ht,0,sizeof(ht));
    memset(fir,0,sizeof(fir));
    memset(sec,0,sizeof(sec));
    memset(pos,0,sizeof(pos));
    memset(tong,0,sizeof(tong));
    copy(s + 1,s + len + 1,sett + 1);
    sort(sett + 1,sett + 1 + len);
    cnt = unique(sett + 1,sett + 1 + len) - sett - 1;
    for (int i = 1; i <= len; i++)
        a[i] = lower_bound(sett + 1,sett + 1 + cnt,s[i]) - sett;
    for (int i = 1; i <= len; i++)
        tong[a[i]]++;
    for (int i = 1; i <= len; i++)
        tong[i] += tong[i - 1];
    for (int i = 1; i <= len; i++)
        rk[i] = tong[a[i] - 1] + 1;
    for (int t = 1; t <= len; t *= 2)
    {
        for (int i = 1; i <= len; i++)
            fir[i] = rk[i];
        for (int i = 1; i <= len; i++)
        {
            if (i + t > len)
                sec[i] = 0;
            else
                sec[i] = rk[i + t];
        }
        fill(tong,tong + 1 + len,0);
        for (int i = 1; i <= len; i++)
            tong[sec[i]]++;
        for (int i = 1; i <= len; i++)
            tong[i] += tong[i - 1];
        for (int i = 1; i <= len; i++)
            pos[len - --tong[sec[i]]] = i;
        fill(tong,tong + 1 + len,0);
        for (int i = 1; i <= len; i++)
            tong[fir[i]]++;
        for (int i = 1; i <= len; i++)
            tong[i] += tong[i - 1];
        for (int i = 1; i <= len; i++)
        {
            int temp = pos[i];
            sa[tong[fir[temp]]--] = temp;
        }
        bool flag = true;
        int last = 0;
        for (int i = 1; i <= len; i++)
        {
            int temp = sa[i];
            if (!last)
                rk[temp] = 1;
            else if (fir[temp] == fir[last] && sec[temp] == sec[last])
            {
                rk[temp] = rk[last];
                flag = false;
            }
            else
                rk[temp] = rk[last] + 1;
            last = temp;
        }
        if (flag)
            break;
    }
    int k = 0;
    for (int i = 1; i <= len; i++)
    {
        if (rk[i] == 1)
            k = 0;
        else
        {
            if (k)
                k--;
            int j = sa[rk[i] - 1];
            while (i + k <= len && j + k <= len && a[i + k] == a[j + k])
                k++;
        }
        ht[rk[i]] = k;
    }
}

int main()
{
    while (~scanf("%s",s1 + 1))
    {
        len = ans = 0;
        len1 = strlen(s1 + 1);
        scanf("%s",s2 + 1);
        len2 = strlen(s2 + 1);
        for (int i = 1; i <= len1; i++)
            s[++len] = s1[i];
        s[++len] = '&';
        poss = len;
        for (int i = 1; i <= len2; i++)
            s[++len] = s2[i];
        solve();
        for (int i = 2; i <= len; i++)
            if (ht[i] > ans && ((sa[i - 1] < poss && sa[i] > poss) || (sa[i - 1] > poss && sa[i] < poss)))
                ans = ht[i];
        printf("%d\n",ans);
    }

    return 0;
}