Poj-3693 maximum repetition substring)

Description

The repetition number of a string is defined as the maximum numberRSuch that the string can be partitionedRSame consecutive substrings. For example, the repetition number of "ababab" is 3 and "Ababa" is 1.

Given a string containing lowercase letters, you are to find a substring of it with maximum repetition number.

Input

The input consists of multiple test cases. Each test case contains exactly one line, which
Gives a non-empty string consisting of lowercase letters. The length of the string will not be greater than 100,000.

The last test case is followed by a line containing '#'.

Output

For each test case, print a line containing the test case number (beginning with 1) followed by the substring of maximum repetition number. if there are multiple substrings of maximum repetition number, print the lexicographically smallest one.

Sample Input

ccabababcdaabbccaa#

Sample output

Case 1: abababCase 2: aa

Question: Give a string and find the continuous repeated substring with the most repetitions.
Idea: worship the suffix array God thesis and cxlove's question: it is easier to understand that the enumerated length is l, and then it can be seen that the string with the length of L appears several times in a row at most.
STR [0], STR [L], STR [2L]… There must be two consecutive strings.
Then enumerate the two consecutive characters, and then match the two characters to see how far it can be matched.
That is to say, it is matched before and after STR [il + L]. Here we query the longest public prefix of suffix (IL) and suffix (IL + l, the rank value can be used to find the ranking between IL and IL + L. We want to query the minimum value of height in this range. Through rmq preprocessing, the query complexity is 0 (1,
If the LCP length is m, the answer is obviously M/L + 1, but this is not necessarily the best, because the beginning and end of the answer are not necessarily at the enumerated position. my solution is to consider the meaning of M % L. We can think that M % L characters are added to the end, but we can think that we have less (L-M % L) characters before! Therefore, we obtain the longest public prefix of the suffix Jl-(L-M % L) and the suffix (J + 1) * l-(L-M % L. That is to put the previous range prefix L-M % L. Then save the length of the maximum value L. Because the question requires the minimum Lexicographic Order, the SA array is used for enumeration. The first group obtained must be the smallest Lexicographic Order.

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#include <queue>using namespace std;const int maxn = 100010;int sa[maxn]; int t1[maxn], t2[maxn], c[maxn];int rank[maxn], height[maxn];void build_sa(int s[], int n, int m) {    int i, j, p, *x = t1, *y = t2;    for (i = 0; i < m; i++) c[i] = 0;    for (i = 0; i < n; i++) c[x[i] = s[i]]++;    for (i = 1; i < m; i++) c[i] += c[i-1];    for (i = n-1; i >= 0; i--) sa[--c[x[i]]] = i;    for (j = 1; j <= n; j <<= 1) {        p = 0;        for (i = n-j; i < n; i++) y[p++] = i;        for (i = 0; i < n; i++)             if (sa[i] >= j)                 y[p++] = sa[i] - j;        for (i = 0; i < m; i++) c[i] = 0;        for (i = 0; i < n; i++) c[x[y[i]]]++;        for (i = 1; i < m; i++) c[i] += c[i-1];        for (i = n-1; i >= 0; i--) sa[--c[x[y[i]]]] = y[i];        swap(x, y);        p = 1, x[sa[0]] = 0;        for (i = 1; i < n; i++)             x[sa[i]] = y[sa[i-1]] == y[sa[i]] && y[sa[i-1]+j] == y[sa[i]+j] ? p-1 : p++;        if (p >= n) break;        m = p;    }}void getHeight(int s[],int n) {    int i, j, k = 0;    for (i = 0; i <= n; i++)        rank[sa[i]] = i;    for (i = 0; i < n; i++) {        if (k) k--;        j = sa[rank[i]-1];        while (s[i+k] == s[j+k]) k++;        height[rank[i]] = k;    }}int dp[maxn][30];char str[maxn];int r[maxn];int a[maxn];void initRMQ(int n) {    int m = floor(log(n+0.0) / log(2.0));      for (int i = 1; i <= n; i++)         dp[i][0] = height[i];      for (int i = 1; i <= m; i++) {          for (int j = n; j; j--) {              dp[j][i] = dp[j][i-1];              if (j+(1<<(i-1)) <= n)                  dp[j][i] = min(dp[j][i], dp[j+(1<<(i-1))][i-1]);          }      }  }int lcp(int l, int r) {    int a = rank[l], b = rank[r];      if (a > b)         swap(a,b);      a++;      int m = floor(log(b-a+1.0) / log(2.0));      return min(dp[a][m], dp[b-(1<<m)+1][m]);  }int main() {    int cas = 1;    while (scanf("%s", str) != EOF && str[0] != ‘#‘) {        int n = strlen(str);        for (int i = 0; i <= n; i++)            r[i] = str[i];        build_sa(r, n+1, 128);        getHeight(r, n);        initRMQ(n);        int cnt = 0, Max = 0;        for (int l = 1; l < n; l++) {            for (int i = 0; i + l < n; i += l) {                int tmp = lcp(i, i+l);                int step = tmp / l + 1;                int k = i - (l - tmp % l);                if (k >= 0 && tmp % l)                     if (lcp(k, k+l) >= tmp)                        step++;                if (step > Max) {                    Max = step;                    cnt = 0;                    a[cnt++] = l;                }                else if (step == Max)                    a[cnt++] = l;            }            }        int len = -1, st;        for (int i = 1; i <= n && len == -1; i++)             for (int j = 0; j < cnt; j++) {                int l = a[j];                if (lcp(sa[i], sa[i]+l) >= (Max - 1) * l) {                    len = l;                    st = sa[i];                    break;                }            }        str[st + len * Max] = 0;        printf("Case %d: %s\n", cas++, str + st);    }    return 0;}

Poj-3693 maximum repetition substring)