meaning of the title
Sol
warning: the following is only 95 points. I took 1w + pictures and found no mistakes. I said that I could do nothing
We consider the ranking of each string to update the answer. Obviously, the prefix of suffix \ (1 \) must be the answer with the smallest dictionary order of the current length, but not necessarily the answer with the smallest left endpoint. Therefore, we need to use a data structure to maintain all feasible left endpoints. Then enumerate all the suffixes to update the answers.
#include<bits/stdc++.h> using namespace std; const int MAXN = 2e6 + 10, SS = 6e5 + 10, INF = 1e9 + 10; template <typename A, typename B> inline bool chmin(A &a, B b){if(a > b) {a = b; return 1;} return 0;} template <typename A, typename B> inline bool chmax(A &a, B b){if(a < b) {a = b; return 1;} return 0;} inline int read() { char c = getchar(); int x = 0, f = 1; while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();} while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar(); return x * f; } int CharSet, N, M, ans[SS]; int s[SS]; char stmp[SS]; int rak[SS], tp[SS], tax[MAXN * 10], sa[SS], height[SS]; void Sort() { for(int i = 0; i <= M; i++) tax[i] = 0; for(int i = 1; i <= N; i++) tax[rak[i]]++; for(int i = 1; i <= M; i++) tax[i] += tax[i - 1]; //for(int i = N; i >= 1; i--) sa[i] = tp[tax[rak[i]]--]; for(int i = N; i >= 1; i--) sa[tax[rak[tp[i]]]--] = tp[i]; } void SuffixArraryBuild() { M = CharSet; for(int i = 1; i <= N; i++) rak[i] = s[i], tp[i] = i; Sort(); for(int w = 1, p; w < N; w <<= 1, M = p) { p = 0; for(int i = N - w + 1; i <= N; i++) tp[++p] = i; for(int i = 1; i <= N; i++) if(sa[i] > w) tp[++p] = sa[i] - w; Sort(); swap(rak, tp); rak[sa[1]] = p = 1; for(int i = 2; i <= N; i++) rak[sa[i]] = (tp[sa[i]] == tp[sa[i - 1]] && tp[sa[i] + w] == tp[sa[i - 1] + w] ? p : ++p); } for(int i = 1, k = 0; i <= N; i++) { int j = sa[rak[i] - 1]; if(k) k--; while(j && s[i + k] == s[j + k]) k++;//mmp didn't write j height[rak[i]] = k;//tag } } #define ls k << 1 #define rs k << 1 | 1 int mn[MAXN]; void update(int k) { mn[k] = min(mn[ls], mn[rs]); } void Build(int k, int l, int r) { if(l == r) {mn[k] = sa[l]; return ;} int mid = l + r >> 1; Build(ls, l, mid); Build(rs, mid + 1, r); update(k); } int IntQuery(int k, int l, int r, int ql, int qr) { if(ql <= l && r <= qr) return mn[k]; int mid = l + r >> 1; if(ql > mid) return IntQuery(rs, mid + 1, r, ql, qr); else if(qr <= mid) return IntQuery(ls, l, mid, ql, qr); else return min(IntQuery(ls, l, mid, ql, qr), IntQuery(rs, mid + 1, r, ql, qr)); } int f[MAXN][20]; int Find(int pos, int cur) {//height from pos, first position < cur if(height[pos] < cur) return pos - 1; for(int i = 19; i >= 1; i--) { while(f[pos][i] >= cur && f[pos][i] <= INF && pos + (1 << i) - 1 <= N) pos = pos + (1 << i) - 1; } return pos; } void solve() { Build(1, 1, N); memset(f, 0x3f, sizeof(f)); for(int i = 1; i <= N; i++) f[i][0] = height[i]; for(int j = 1; j <= 19; j++) for(int i = 1; i + (1 << j) - 1<= N; i++) chmin(f[i][j], min(f[i][j - 1], f[i + (1 << j - 1)][j - 1])); int cur = 1;//The answer we're looking for for(int rk = 1; rk <= N; rk++) { if((N - sa[rk] + 1) < cur) continue; int pos = sa[rk]; for(int i = pos + cur - 1; i <= N; i++) { int j = Find(rk + 1, cur); ans[cur++] = IntQuery(1, 1, N, rk, j); } } } signed main() { CharSet = read(); N = read(); if(CharSet == 26) { scanf("%s", stmp + 1); for(int i = 1; i <= N; i++) s[i] = stmp[i] - 'a'; } else for(int i = 1; i <= N; i++) s[i] = read(); SuffixArraryBuild(); solve(); for(int i = 1; i <= N; i++) cout << ans[i] << ' '; return 0; }