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人生第一道后缀数组的题目。首先要对输入的串进行差分处理，差分后长度为（\(n - 1\)）的相同子段就是原串中长度为\(n\)的相同（可变调）子段。求出来\(height\)以后，二分一个最大不相交重复子段长度，在\(can\_use\)里面维护长度为\(k\)的段划分，判断不相交只要用\(sa\)的差值算一下长度就好。
\(p.s.\)细节真的很多。
#include <bits/stdc++.h>
using namespace std;
const int N = 5010;
int n, m = 255, s[N], sa[N], rk[N], tp[N], _rk[N], bin[N], height[N];
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 () {
for (int i = 1; i <= n; ++i) {
rk[i] = s[i];
tp[i] = 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;
}
}
void get_height () {
int k = 0;
for (int i = 1; i <= n; ++i) {
if (k != 0) k = k - 1;
int j = sa[rk[i] - 1];
while (s[i + k] == s[j + k]) {
k = k + 1;
}
height[rk[i]] = k;
}
}
const int INF = 0x3f3f3f3f;
bool can_use (int k) {
//是否有长度>=k的不交叉子串
int max_sa = sa[1], min_sa = sa[1];
for (int i = 2; i <= n; ++i) {
if (height[i] >= (k - 1)) {
//即原串中旋律长度 >= k
max_sa = max (max_sa, sa[i]);
min_sa = min (min_sa, sa[i]);
if (max_sa - min_sa >= k) {
return true;
}
} else {
max_sa = sa[i];
min_sa = sa[i];
}
}
return false;
}
int main () {
cin >> n;
for (int i = 1; i <= n; ++i) {
cin >> s[i];
}
for (int i = n; i >= 1; --i) {
s[i] = s[i] - s[i - 1] + 90;
//把s差分,出现n-1个相同项说明最长为n
}
// printf ("str Test : "); for (int i = 1; i <= n; ++i) printf ("%3d ", s[i]); putchar ('\n');
suffix_sort ();
get_height ();
// printf ("height Test : "); for (int i = 1; i <= n; ++i) printf ("%3d ", height[i]); printf ("\n");
int l = 0, r = n;
while (l < r) {
// printf ("l = %d, r = %d\n", l, r);
int mid = (l + r + 1) >> 1;
if (can_use (mid)) {
//存在长度 >= mid的不交叉子串
l = mid;
} else {
r = mid - 1;
}
}
if (l < 5) printf ("0\n");
else printf ("%d\n", l);
}