SPOJ - REPEATS —— 后缀数组重复次数最多的连续重复子串 - 开发语言

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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);

     }

 }