題面

題目傳送門

解法

后綴數組模板題吧……

將字符串兩倍，然后求一遍sa數組即可

時間復雜度：\(O(n\ log\ n)\)

代碼

#include <bits/stdc++.h> #define N 200010 using namespace std; char st[N]; int len; struct SuffixArray {int x[N], y[N], sa[N], cnt[N];void getsa() {int n = strlen(st + 1), m = 200;for (int i = 1; i <= n; i++) st[i + n] = st[i]; n *= 2;for (int i = 1; i <= n; i++) cnt[x[i] = st[i]]++;for (int i = 2; i <= m; i++) cnt[i] += cnt[i - 1];for (int i = n; i; i--) sa[cnt[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++) cnt[i] = 0;for (int i = 1; i <= n; i++) cnt[x[i]]++;for (int i = 2; i <= m; i++) cnt[i] += cnt[i - 1];for (int i = n; i; i--) sa[cnt[x[y[i]]]--] = y[i], y[i] = 0;swap(x, y); x[sa[1]] = 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 (n == num) break; m = num;}}void solve() {int n = strlen(st + 1);for (int i = 1; i <= n; i++) {int x = sa[i];if (x <= len) cout << st[x + len - 1];}cout << "\n";} } SA; int main() {scanf(" %s", st + 1);len = strlen(st + 1);SA.getsa(); SA.solve(); return 0; }

轉載于:https://www.cnblogs.com/copperoxide/p/9478372.html

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