PAT (Top Level) Practice1005 Programming Pattern （35 分）

题目链接：https://pintia.cn/problem-sets/994805148990160896/problems/994805154748940288

用后缀数组求出长度为n所有子字符串的大小关系，然后得出重复最多的。

算法：后缀数组

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<set>
#define maxn 1100000
using namespace std;
int n;
int sa[maxn], ranks[maxn];
int height[maxn];
int t1[maxn], t2[maxn];
int c[maxn];
string s;
int m;
bool cmp(int *r, int a, int b, int l) {return r[a] == r[b] && r[a + l] == r[b + l];
}
void da(int t) {int p;m = s.length();int *x = t1, *y = t2;for (int i = 0; i < t; i++) {c[i] = 0;}for (int i = 0; i < m; i++) {c[x[i] = s[i]]++;}for (int i = 1; i < t; i++) {c[i] += c[i - 1];}for (int i = m - 1; i >= 0; i--) {sa[--c[x[i]]] = i;}for (int k = 1; k <= m;) {p = 0;for (int i = m - k; i < m; i++) {y[p++] = i;}for (int i = 0; i < m; i++) {if (sa[i] >= k) {y[p++] = sa[i] - k;}}for (int i = 0; i < t; i++) {c[i] = 0;}for (int i = 0; i < m; i++) {c[x[y[i]]]++;}for (int i = 1; i < t; i++) {c[i] += c[i - 1];}for (int i = m - 1; i >= 0; i--)sa[--c[x[y[i]]]] = y[i];swap(x, y);p = 1;x[sa[0]] = 0;for (int i = 1; i < m; i++) {x[sa[i]] = cmp(y, sa[i - 1], sa[i], k) ? p - 1 : p++;}if (p >= m)break;t = p;k = k << 1;if (k >= n){for (int i = 0; i < m; i++) {ranks[sa[i]] = i;}break;}}
}
void getheight() {int k = 0;for (int i = 0; i < m; height[ranks[i++]] = k) {if (k)--k;if (ranks[i] == 0) {height[ranks[i]] = 0;continue;}for (int j = sa[ranks[i] - 1]; s[i + k] == s[j + k];) {k++;if (k >= n) {k = n; break;                              }}}
}
int main() {cin >> n;cin.get();getline(cin, s);da(128);getheight();int h = sa[0], _h = -1;int ans = 1, ans1 = 1;for (int i = 1; i < m; i++) {if (height[i] >= n) {ans++;}else {if (ans > ans1||(_h==-1)||(ans==ans1&&s[h]<s[_h])) {ans1 = ans;_h = h;}ans = 1, h = sa[i];}}if (ans > ans1 || (ans == ans1 && s[h] < s[_h])) {ans1 = ans; _h = h;}cout << s.substr(_h, n) << " " << ans1 << endl;return 0;
}

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