Description

题库链接

给出一个长度为 \(n\) 的字符串，求重复次数最多的连续重复子串。

\(1\leq n\leq 50000\)

Solution

Code

#include <bits/stdc++.h>
#define log2 LOG
using namespace std;
const int N = 100000+5, inf = ~0u>>1;char ch[N];
int n, m, x[N<<1], y[N<<1], c[N], sa[N], rk[N], height[N], Case;
int log2[N], bin[30], f[30][N], ans, t;void get() {for (int i = 1; i <= m; i++) c[i] = 0;for (int i = 1; i <= n; i++) c[x[i] = ch[i]]++;for (int i = 2; i <= m; i++) c[i] += c[i-1];for (int i = n; i >= 1; i--) sa[c[x[i]]--] = i;for (int k = 1; k <= n; k <<= 1) {int num = 0;for (int i = n-k+1; i <= n; i++) y[++num] = i;for (int i = 1; i <= n; i++) if (sa[i] > k) y[++num] = sa[i]-k;for (int i = 1; i <= m; i++) c[i] = 0;for (int i = 1; i <= n; i++) c[x[i]]++;for (int i = 2; i <= m; i++) c[i] += c[i-1];for (int i = n; i >= 1; i--) sa[c[x[y[i]]]--] = y[i];swap(x, y); x[sa[1]] = num = 1;for (int i = 2; i <= n; i++)x[sa[i]] = (y[sa[i]] == y[sa[i-1]] && y[sa[i]+k] == y[sa[i-1]+k]) ? num : ++num;if ((m = num) == n) break;}for (int i = 1; i <= n; i++) rk[sa[i]] = i;for (int i = 1, k = 0; i <= n; i++) {if (rk[i] == 1) continue;if (k) --k; int j = sa[rk[i]-1];while (j+k <= n && i+k <= n && ch[i+k] == ch[j+k]) ++k;height[rk[i]] = k;}
}
void rmq() {int t = log2[n];for (int i = 1; i <= n; i++) f[0][i] = height[i];for (int i = 1; i <= t; i++)for (int j = 1; j+bin[i]-1 <= n; j++)f[i][j] = min(f[i-1][j], f[i-1][j+bin[i-1]]);
}
int query(int a, int b) {a = rk[a], b = rk[b];if (a > b) swap(a, b); ++a;int t = log2[b-a+1];return min(f[t][a], f[t][b-bin[t]+1]);
}
void work() {bin[0] = 1; log2[0] = -1;for (int i = 1; i <= 25; i++) bin[i] = (bin[i-1]<<1);for (int i = 1; i < N; i++) log2[i] = log2[i>>1]+1;scanf("%d", &t);while (t--) {scanf("%d", &n); getchar(); m = 255; ans = 0;for (int i = 1; i <= n; i++) scanf("%c", &ch[i]), getchar();get(); rmq();for (int l = 1; l <= n; l++)for (int i = 1; i+l <= n; i += l) {int k = query(i, i+l), t = l-k%l, p = i-t, m = k/l+1;if (p > 0 && query(p+l, p) >= l-k%l) ++m;if (m > ans) ans = m;}printf("%d\n", ans);}
}
int main() {work(); return 0; }