SPOJ - REPEATS Repeats (后缀数组+RMQ)

题意：求一个串中出现重复子串次数最多的数目。

析：枚举每个长度的子串，至少要重复两次，必然会经过s[l*i]中相邻的两个，然后再分别向前和向后匹配即可。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")

#include <cstdio>

#include <string>

#include <cstdlib>

#include <cmath>

#include <iostream>

#include <cstring>

#include <set>

#include <queue>

#include <algorithm>

#include <vector>

#include <map>

#include <cctype>

#include <cmath>

#include <stack>

#include <sstream>

#define debug() puts("++++");

#define gcd(a, b) __gcd(a, b)

#define lson l,m,rt<<1

#define rson m+1,r,rt<<1|1

#define freopenr freopen("in.txt", "r", stdin)

#define freopenw freopen("out.txt", "w", stdout)

using namespace std;

typedef long long LL;

typedef unsigned long long ULL;

typedef pair<int, int> P;

const int INF = 0x3f3f3f3f;

const LL LNF = 1e16;

const double inf = 0x3f3f3f3f3f3f;

const double PI = acos(-1.0);

const double eps = 1e-8;

const int maxn = 50000 + 10;

const int mod = 1e9 + 7;

const int dr[] = {-1, 0, 1, 0};

const int dc[] = {0, 1, 0, -1};

const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};

int n, m;

const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

inline bool is_in(int r, int c){

  return r >= 0 && r < n && c >= 0 && c < m;

}

struct Array{

  int sa[maxn], s[maxn], t[maxn], t2[maxn];

  int r[maxn], h[maxn], c[maxn], dp[maxn][20];

  int n;

  void init(){

    n = 0;  memset(sa, 0, sizeof sa);

  }

  void build_sa(int m){

    int *x = t, *y = t2;

    for(int i = 0; i < m; ++i)  c[i] = 0;

    for(int i = 0; i < n; ++i)  ++c[x[i] = s[i]];

    for(int i = 1; i < m; ++i)  c[i] += c[i-1];

    for(int i = n-1; i >= 0; --i)  sa[--c[x[i]]] = i;

    for(int k = 1; k <= n; k <<= 1){

      int p = 0;

      for(int i = n-k; i < n; ++i)  y[p++] = i;

      for(int i = 0; i < n; ++i)  if(sa[i] >= k)  y[p++] = sa[i] - k;

      for(int i = 0; i < m; ++i)  c[i] = 0;

      for(int i = 0; i < n; ++i)  ++c[x[y[i]]];

      for(int i = 1; i < m; ++i)  c[i] += c[i-1];

      for(int i = n-1; i >= 0; --i)  sa[--c[x[y[i]]]] = y[i];

      swap(x, y);

      p = 1;  x[sa[0]] = 0;

      for(int i = 1; i < n; ++i)

        x[sa[i]] = y[sa[i-1]] == y[sa[i]] && y[sa[i-1]+k] == y[sa[i]+k] ? p-1 : p++;

      if(p >= n)  break;

      m = p;

    }

  }

  void getHight(){

    int k = 0;

    for(int i = 0; i < n; ++i)  r[sa[i]] = i;

    for(int i = 0; i < n; ++i){

      if(k)  --k;

      int j = sa[r[i]-1];

      while(s[i+k] == s[j+k])  ++k;

      h[r[i]] = k;

    }

  }

  void rmq_init(){

    for(int i = 1; i <= n; ++i)  dp[i][0] = h[i];

    for(int j = 1; (1<<j) <= n; ++j)

      for(int i = 1; i + (1<<j) <= n; ++i)

        dp[i][j] = min(dp[i][j-1], dp[i+(1<<j-1)][j-1]);

  }

  int query(int L, int R){

    L = r[L];  R = r[R];

    if(L > R)  swap(L, R);

    ++L;

    int k = log(R-L+1) / log(2.0);

    return min(dp[L][k], dp[R-(1<<k)+1][k]);

  }

};

Array arr;

char s[5];

int main(){

  int T;  cin >> T;

  while(T--){

    scanf("%d", &n);

    arr.init();

    for(int i = 0; i < n; ++i){

      scanf("%s", s);

      arr.s[arr.n++] = s[0] - 'a' + 1;

    }

    arr.s[arr.n++] =  0;

    arr.build_sa(5);

    arr.getHight();

    arr.rmq_init();

    int ans = 0;

    for(int i = 1; i <= n; ++i)

      for(int j = 0; j + i <= n; j += i){

        int k = arr.query(j, j + i);

        int res = k / i + 1;

        int t = j - (i - k % i);

        if(t >= 0 && arr.query(t, t + i) >= k) ++res;

        ans = max(ans, res);

      }

    printf("%d\n", ans);

  }

  return 0;

}