思路:求第k大字串,求出sam上每个节点开始能识别多少字串,然后从起点开始跑就好啦。
#include<bits/stdc++.h> #define LL long long #define fi first #define se second #define mk make_pair #define PII pair<int, int> #define PLI pair<LL, int> #define PDD pair<double,double> #define ull unsigned long long using namespace std; const int N = 90000 + 7; const int inf = 0x3f3f3f3f; const LL INF = 0x3f3f3f3f3f3f3f3f; const int mod = 1e9 + 7; const double eps = 1e-8; int n, ans[N]; LL dp[N << 1]; char s[N]; struct SuffixAutomaton { int last, cur, cnt, ch[N<<1][26], id[N<<1], fa[N<<1], dis[N<<1], sz[N<<1], c[N]; SuffixAutomaton() {cur = cnt = 1;} void init() { for(int i = 1; i <= cnt; i++) { memset(ch[i], 0, sizeof(ch[i])); sz[i] = c[i] = dis[i] = fa[i] = 0; } cur = cnt = 1; } void extend(int c, int id) { last = cur; cur = ++cnt; int p = last; dis[cur] = id; for(; p && !ch[p][c]; p = fa[p]) ch[p][c] = cur; if(!p) fa[cur] = 1; else { int q = ch[p][c]; if(dis[q] == dis[p]+1) fa[cur] = q; else { int nt = ++cnt; dis[nt] = dis[p]+1; memcpy(ch[nt], ch[q], sizeof(ch[q])); fa[nt] = fa[q]; fa[q] = fa[cur] = nt; for(; ch[p][c]==q; p=fa[p]) ch[p][c] = nt; } } sz[cur] = 1; } void getSize(int n) { for(int i = 1; i <= cnt; i++) c[dis[i]]++; for(int i = 1; i <= n; i++) c[i] += c[i-1]; for(int i = cnt; i >= 1; i--) id[c[dis[i]]--] = i; for(int i = cnt; i >= 1; i--) { int p = id[i]; sz[fa[p]] += sz[p]; } } void work(int u, int k) { if(!k) return; LL ret = 0; for(int i = 0; i < 26; i++) { if(!ch[u][i]) continue; if(ret + dp[ch[u][i]] + 1 >= k) { putchar(i + 'a'); work(ch[u][i], k - ret - 1); break; } ret += dp[ch[u][i]] + 1; } } void solve() { for(int i = cnt; i >= 1; i--) { int u = id[i]; dp[u] = 0; for(int c = 0; c < 26; c++) if(ch[u][c]) dp[u] += dp[ch[u][c]] + 1; } int q, k; scanf("%d", &q); while(q--) { scanf("%d", &k); work(1, k); puts(""); } } } sam; int main() { scanf("%s", s + 1); n = strlen(s + 1); for(int i = 1; i <= n; i++) sam.extend(s[i]-'a', i), ans[i] = 0; sam.getSize(n); sam.solve(); return 0; } /* */