SP687 REPEATS - Repeats
给定字符串,求重复次数最多的连续重复子串。
题目很简单,被细节坑惨了。。。
前置的一个推论:请看这里。
#include <bits/stdc++.h>
using namespace std;
const int N = 50010;
struct String {
char s[N]; int st[N][17];
int n, m, sa[N], tp[N], rk[N], _rk[N], bin[N], height[N];
void clear () {
memset (s, 0, sizeof (s));
memset (sa, 0, sizeof (sa));
memset (tp, 0, sizeof (tp));
memset (rk, 0, sizeof (rk));
memset (st, 0, sizeof (st));
memset (_rk, 0, sizeof (_rk));
memset (bin, 0, sizeof (bin));
memset (height, 0, sizeof (height));
}
void base_sort () {
for (int i = 0; i <= m; ++i) bin[i] = 0;
for (int i = 1; i <= n; ++i) bin[rk[tp[i]]]++;
for (int i = 1; i <= m; ++i) bin[i] += bin[i - 1];
for (int i = n; i >= 1; --i) sa[bin[rk[tp[i]]]--] = tp[i];
}
void suffix_sort () {
m = 255; ;
for (int i = 1; i <= n; ++i) {
tp[i] = i, rk[i] = s[i];
}
base_sort ();
for (int w = 1; w <= n; w <<= 1) {
int cnt = 0;
for (int i = n - w + 1; i <= n; ++i) tp[++cnt] = i;
for (int i = 1; i <= n; ++i) if (sa[i] > w) tp[++cnt] = sa[i] - w;
base_sort ();
memcpy (_rk, rk, sizeof (rk));
rk[sa[1]] = cnt = 1;
for (int i = 2; i <= n; ++i) {
rk[sa[i]] = _rk[sa[i]] == _rk[sa[i - 1]] && _rk[sa[i] + w] == _rk[sa[i - 1] + w] ? cnt : ++cnt;
}
if (cnt == n) break;
m = cnt;
}
int k = 0;
for (int i = 1; i <= n; ++i) {
if (k != 0) --k;
int j = sa[rk[i] - 1];
while (s[i + k] == s[j + k]) ++k;
height[rk[i]] = k;
}
int mx = log2 (n);
for (int i = 1; i <= n; ++i) {
st[i][0] = height[i];
}
for (int i = 1; i <= mx; ++i) {
for (int j = 1; j + (1 << i) - 1 <= n; ++j) {
st[j][i] = min (st[j][i - 1], st[j + (1 << (i - 1))][i - 1]);
}
}
}
int lcp (int l, int r) {
if (l == r) return n - l + 1;
l = rk[l], r = rk[r];
if (l > r) swap (l, r); ++l;
if (l > r) return 0;
int mx = log2 (r - l + 1);
return min (st[l][mx], st[r - (1 << mx) + 1][mx]);
}
}s1, s2;
int T, n; char s[N];
int main () {
cin >> T;
while (T--) {
int ans = 0;
cin >> n; s1.n = s2.n = n;
s1.clear (); s2.clear ();
for (int i = 1; i <= n; ++i) {
cin >> s[i];
s1.s[i] = s[i];
s2.s[n - i + 1] = s[i];
}
s1.suffix_sort ();
s2.suffix_sort ();
for (int len = 1; len <= n; ++len) {//枚举重复子串的长度
for (int p = 1; p + len <= n; p += len) {
int K = s1.lcp (p, p + len) + s2.lcp (n - (p) + 1, n - (p + len) + 1) - 1;
ans = max (ans, K / len + 1);
}
}
cout << max (ans, 1) << endl;
}
}