SPOJ

题目链接: SPOJ - REPEATS

题目大意

求出现次数最多的连续重复子串, 输出出现的次数

思路

穷举长度len, 如果长度为len的子串出现了两次, 记这个子串所有重复连在为s, 原来的串为r, 那么s肯定包括了r[0], r[len], r[len*2], r[len*3]…中相邻两个, 所以只须看字符r[len*i]和r[len*(i+1)]往前和往后各能匹配到多远，记这个总长度为K，那么这里连续出现了K/L+1次。
如图:

向后能匹配多远可以利用高度数组直接求出, 向前也可以将原串翻转再求一个高度数组得到, 但还有更简洁的方法: 题解

代码

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxn = 5e4 + 100;int n;int mi[maxn][30];void init_rmq(int a[], int n){    for (int i = 0; i < n; ++i) mi[i][0] = a[i];    for (int j = 1; (1 << j) <= n; ++j)    {        for (int i = 0; i + (1 << j) <= n; ++i)        {            mi[i][j] = min(mi[i][j - 1], mi[i + (1 << (j - 1))][j - 1]);        }    }}int rmq(int l, int r){    int k = 0;    while ((1 << (k + 1)) <= r - l + 1) ++k;    return min(mi[l][k], mi[r - (1 << k) + 1][k]);}int rk[maxn], tmp[maxn], sa[maxn], k, lcp[maxn];bool cmp_sa(int i, int j){    if (rk[i] != rk[j]) return rk[i] < rk[j];    return (i + k <= n ? rk[i + k] : -1) < (j + k <= n ? rk[j + k] : -1);}void construct_sa(char *s){    for (int i = 0; i <= n; ++i)    {        sa[i] = i;        rk[i] = s[i];    }    rk[n] = -1;    for (k = 1; k <= n; k *= 2)    {        sort(sa, sa + n + 1, cmp_sa);        tmp[sa[0]] = 0;        for (int i = 1; i <= n; ++i)            tmp[sa[i]] = tmp[sa[i - 1]] + (cmp_sa(sa[i - 1], sa[i]) ? 1 : 0);        copy(tmp, tmp + n + 1, rk);    }}void construct_lcp(char *s){    for (int i = 0; i <= n; ++i) rk[sa[i]] = i;    int h = 0;    lcp[0] = 0;    for (int i = 0; i < n; ++i)    {        int j = sa[rk[i] - 1];        if (h > 0) --h;        for (; j + h < n && i + h < n; ++h)            if (s[j + h] != s[i + h]) break;        lcp[rk[i] - 1] = h;    }}int query(int a, int b){    a = rk[a];    b = rk[b];    if (a > b) swap(a, b);    return rmq(a, b - 1);}char s[maxn];int main(){    int T;    for (scanf("%d", &T); T; --T)    {        scanf("%d", &n);        char ts[10];        for (int i = 0; i < n; ++i)        {            scanf("%s", ts);            s[i] = ts[0];        }        s[n] = '\0';        construct_sa(s);        construct_lcp(s);        init_rmq(lcp, n + 1);        int ans = 0;        for (int len = 1; len <= n; ++len)        {            for (int i = 0; i + len < n; i += len)            {                int t = query(i, i + len);//向后匹配长度                //后面的匹配有t/len次重复, 还有一段长度为t%len的不完整, (len-t%len)为不完整子串缺少的长度                int p = i - (len - t % len);                t = t / len + 1;                if (p >= 0 && query(p, p + len) >= len) ++t;//如果前面还有一段重复, ++t                ans = max(ans, t);            }        }        cout << ans << endl;    }    return 0;}