SPOJ687---REPEATS - Repeats（后缀数组+RMQ）

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.

方法还是论文上说的方法，我用2个后缀数组，另外一个是字符串反转求出来的，然后分别LCP

/*************************************************************************
    > File Name: SPOJ-687.cpp
    > Author: ALex
    > Mail: zchao1995@gmail.com 
    > Created Time: 2015年04月07日 星期二 18时16分49秒
 ************************************************************************/

#include <functional>
#include <algorithm>
#include <iostream>
#include <fstream>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <queue>
#include <stack>
#include <map>
#include <bitset>
#include <set>
#include <vector>

using namespace std;

const double pi = acos(-1.0);
const int inf = 0x3f3f3f3f;
const double eps = 1e-15;
typedef long long LL;
typedef pair <int, int> PLL;

class SuffixArray
{
    public:
        static const int N = 52000;
        int init[N];
        int X[N];
        int Y[N];
        int Rank[N];
        int sa[N];
        int height[N];
        int buc[N];
        int LOG[N];
        int dp[N][20];
        int size;

        void clear()
        {
            size = 0;
        }

        void insert(int n)
        {
            init[size++] = n;
        }

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

        void getsa(int m = 256) //m一般为最大值+1
        {
            init[size] = 0;
            int l, p, *x = X, *y = Y, n = size + 1;
            for (int i = 0; i < m; ++i)
            {
                buc[i] = 0;
            }
            for (int i = 0; i < n; ++i)
            {
                ++buc[x[i] = init[i]];
            }
            for (int i = 1; i < m; ++i)
            {
                buc[i] += buc[i - 1];
            }
            for (int i = n - 1; i >= 0; --i)
            {
                sa[--buc[x[i]]] = i;
            }
            for (l = 1, p = 1; l <= n && p < n; m = p, l *= 2)
            {
                p = 0;
                for (int i = n - l; i < n; ++i)
                {
                    y[p++] = i;
                }
                for (int i = 0; i < n; ++i)
                {
                    if (sa[i] >= l)
                    {
                        y[p++] = sa[i] - l;
                    }
                }
                for (int i = 0; i < m; ++i)
                {
                    buc[i] = 0;
                }
                for (int i = 0; i < n; ++i)
                {
                    ++buc[x[y[i]]];
                }
                for (int i = 1; i < m; ++i)
                {
                    buc[i] += buc[i - 1];
                }
                for (int i = n - 1; i >= 0; --i)
                {
                    sa[--buc[x[y[i]]]] = y[i];
                }
                int i;

                for (swap(x, y), x[sa[0]] = 0, p = 1, i = 1; i < n; ++i)
                { 
                    x[sa[i]] = cmp(y, sa[i - 1], sa[i], l) ? p - 1 : p++; 
                }
            }
        }

        void getheight()
        {
            int h = 0, n = size;
            for (int i = 0; i <= n; ++i)
            {
                Rank[sa[i]] = i;
            }
            height[0] = 0;
            for (int i = 0; i < n; ++i)
            {
                if (h > 0)
                {
                    --h;
                }
                int j =sa[Rank[i] - 1];
                for (; i + h < n && j + h < n && init[i + h] == init[j + h]; ++h);
                height[Rank[i] - 1] = h;
            }
        }   

        //预处理每一个数字的对数，用于rmq，常数优化
        void initLOG()
        {
            LOG[0] = -1;
            for (int i = 1; i < N; ++i)
            {
                LOG[i] = (i & (i - 1)) ? LOG[i - 1] : LOG[i - 1] + 1;
            }
        }

        void initRMQ()
        {
            initLOG();
            int n = size;
            int limit;
            for (int i = 0; i < n; ++i)
            {
                dp[i][0] = height[i];
            }
            for (int j = 1; j <= LOG[n]; ++j)
            {
                limit = n - (1 << j);
                for (int i = 0; i <= limit; ++i)
                {
                    dp[i][j] = min(dp[i][j - 1], dp[i + (1 << (j - 1))][j - 1]);
                }
            }
        }

        int LCP(int a, int b)
        {
            int t;
            a = Rank[a];
            b = Rank[b];
            if (a > b)
            {
                swap(a, b);
            }
            --b;
            t = LOG[b - a + 1];
            return min(dp[a][t], dp[b - (1 << t) + 1][t]);
        }
}SA1, SA2;

char buf[52000];
char str[10];

int main()
{
    int t;
    scanf("%d", &t);
    while (t--)
    {
        SA1.clear();
        SA2.clear();
        int n;
        scanf("%d", &n);
        for (int i = 0; i < n; ++i)
        {
            scanf("%s", str);
            SA1.insert((int)str[0]);
            buf[i] = str[0];
        }
        buf[n] = ‘\0‘;
        for (int i = n - 1; i >= 0; --i)
        {
            SA2.insert((int)buf[i]);
        }
        SA1.getsa((int)(‘b‘) + 1);
        SA2.getsa((int)(‘b‘) + 1);
        SA1.getheight();
        SA2.getheight();
        SA1.initRMQ();
        SA2.initRMQ();
        int ans = 1, s, ans_L = 0;
        for (int L = 1; L <= n; ++L)
        {
            for (int i = 0; i < n; ++i)
            {
                int a = L * i;
                int b = L * (i + 1);
                if (b >= n)
                {
                    break;
                }
                int l1 = SA1.LCP(a, b);
                int l2 = SA2.LCP(n - a - 1, n - b - 1);
                int k = l1 + l2;
                if (k > 0)
                {
                    --k;
                }
                int cnt = k / L + 1;
                if (cnt > ans)
                {
                    ans = cnt;
                    s = a;
                    ans_L = L;
                } 

            }
        }
        printf("%d\n", ans);
    }
    return 0;
}