SPOJ - REPEATS —— 后缀数组 重复次数最多的连续重复子串

题目链接：https://vjudge.net/problem/SPOJ-REPEATS

REPEATS - Repeats

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A string s is called an (k,l)-repeat if s is obtained by concatenating k>=1 times some seed string t with length l>=1. For example, the string

s = abaabaabaaba

is a (4,3)-repeat with t = aba as its seed string. That is, the seed string t is 3 characters long, and the whole string s is obtained by repeating t 4 times.

Write a program for the following task: Your program is given a long string u consisting of characters ‘a’ and/or ‘b’ as input. Your program must find some (k,l)-repeat that occurs as substring within u with k as large as possible. For example, the input string

u = babbabaabaabaabab

contains the underlined (4,3)-repeat s starting at position 5. Since u contains no other contiguous substring with more than 4 repeats, your program must output the maximum k.

Input

In the first line of the input contains H- the number of test cases (H <= 20). H test cases follow. First line of each test cases is n - length of the input string (n <= 50000), The next n lines contain the input string, one character (either ‘a’ or ‘b’) per line, in order.

Output

For each test cases, you should write exactly one interger k in a line - the repeat count that is maximized.

Example

Input:
1
17
b
a
b
b
a
b
a
a
b
a
a
b
a
a
b
a
b

Output:
4

since a (4, 3)-repeat is found starting at the 5th character of the input string.

题意：

给出一个字符串，求该字符串的重复次数最多的连续重复子串，输出重复次数。

题解：

论文上面的题。

代码如下：

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 #include <vector>
 #include <cmath>
 #include <queue>
 #include <stack>
 #include <map>
 #include <string>
 #include <set>
 using namespace std;
 typedef long long LL;
 const int INF = 2e9;
 const LL LNF = 9e18;
 const int MOD = 1e9+;
 const int MAXN = 5e4+;

 bool cmp(int *r, int a, int b, int l)
 {
     return r[a]==r[b] && r[a+l]==r[b+l];
 }

 int r[MAXN], sa[MAXN], Rank[MAXN], height[MAXN];
 int t1[MAXN], t2[MAXN], c[MAXN];
 void DA(int str[], int sa[], int Rank[], int height[], int n, int m)
 {
     n++;
     int i, j, p, *x = t1, *y = t2;
     for(i = ; i<m; i++) c[i] = ;
     for(i = ; i<n; i++) c[x[i] = str[i]]++;
     for(i = ; i<m; i++) c[i] += c[i-];
     for(i = n-; i>=; i--) sa[--c[x[i]]] = i;
     for(j = ; j<=n; j <<= )
     {
         p = ;
         for(i = n-j; i<n; i++) y[p++] = i;
         for(i = ; i<n; i++) if(sa[i]>=j) y[p++] = sa[i]-j;

         for(i = ; i<m; i++) c[i] = ;
         for(i = ; i<n; i++) c[x[y[i]]]++;
         for(i = ; i<m; i++) c[i] += c[i-];
         for(i = n-; i>=; i--) sa[--c[x[y[i]]]] = y[i];

         swap(x, y);
         p = ; x[sa[]] = ;
         for(i = ; i<n; i++)
             x[sa[i]] = cmp(y, sa[i-], sa[i], j)?p-:p++;

         if(p>=n) break;
         m = p;
     }

     int k = ;
     n--;
     for(i = ; i<=n; i++) Rank[sa[i]] = i;
     for(i = ; i<n; i++)
     {
         if(k) k--;
         j = sa[Rank[i]-];
         while(str[i+k]==str[j+k]) k++;
         height[Rank[i]] = k;
     }
 }

 int dp[MAXN][], mm[MAXN];
 void initRMQ(int n, int b[])
 {
     mm[] = -;
     for(int i = ; i<=n; i++)
         dp[i][] = b[i], mm[i] = ((i&(i-))==)?mm[i-]+:mm[i-];
     for(int j = ; j<=mm[n]; j++)
     for(int i = ; i+(<<j)-<=n; i++)
         dp[i][j] = min(dp[i][j-], dp[i+(<<(j-))][j-]);
 }

 int RMQ(int x, int y)
 {
     if(x>y) swap(x, y);
     x++;
     int k = mm[y-x+];
     return min(dp[x][k], dp[y-(<<k)+][k]);
 }

 int main()
 {
     int T, n;
     scanf("%d", &T);
     while(T--)
     {
         scanf("%d", &n);
         for(int i = ; i<n; i++)
         {
             char ch;
             getchar();
             scanf("%c", &ch);
             r[i] = ch-'a'+;
         }
         r[n] = ;
         DA(r, sa, Rank, height, n, );
         initRMQ(n, height);

         int times = , L, R;
         for(int len = ; len<=n; len++)
         for(int pos = ; pos+len<n; pos += len)
         {
             int LCP = RMQ(Rank[pos], Rank[pos+len]);
             int supplement = len - LCP%len;
             int k = pos - supplement;
             if(k>= && LCP%len && RMQ(Rank[k],Rank[k+len])>=supplement)
                 LCP += supplement;
             times = max(times, LCP/len+);
         }
         printf("%d\n", times);
     }
 }