SPOJ - SUBLEX 后缀自动机

SPOJ - SUBLEX

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 =  + ;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + ;
const double eps = 1e-;
 
int n, ans[N];
LL dp[N << ];
char s[N];
 
struct SuffixAutomaton {
    int last, cur, cnt, ch[N<<][], id[N<<], fa[N<<], dis[N<<], sz[N<<], c[N];
    SuffixAutomaton() {cur = cnt = ;}
    void init() {
        for(int i = ; i <= cnt; i++) {
            memset(ch[i], , sizeof(ch[i]));
            sz[i] = c[i] = dis[i] = fa[i] = ;
        }
        cur = cnt = ;
    }
    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] = ;
        else {
            int q = ch[p][c];
            if(dis[q] == dis[p]+) fa[cur] = q;
            else {
                int nt = ++cnt; dis[nt] = dis[p]+;
                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] = ;
    }
    void getSize(int n) {
        for(int i = ; i <= cnt; i++) c[dis[i]]++;
        for(int i = ; i <= n; i++) c[i] += c[i-];
        for(int i = cnt; i >= ; i--) id[c[dis[i]]--] = i;
        for(int i = cnt; i >= ; i--) {
            int p = id[i];
            sz[fa[p]] += sz[p];
        }
    }
    void work(int u, int k) {
        if(!k) return;
        LL ret = ;
        for(int i = ; i < ; i++) {
            if(!ch[u][i]) continue;
            if(ret + dp[ch[u][i]] +  >= k) {
                putchar(i + 'a');
                work(ch[u][i], k - ret - );
                break;
            }
            ret += dp[ch[u][i]] + ;
        }
    }
    void solve() {
        for(int i = cnt; i >= ; i--) {
            int u = id[i]; dp[u] = ;
            for(int c = ; c < ; c++)
                if(ch[u][c]) dp[u] += dp[ch[u][c]] + ;
        }
        int q, k; scanf("%d", &q);
        while(q--) {
            scanf("%d", &k);
            work(, k);
            puts("");
        }
    }
} sam;
 
int main() {
    scanf("%s", s + );
    n = strlen(s + );
    for(int i = ; i <= n; i++)
        sam.extend(s[i]-'a', i), ans[i] = ;
    sam.getSize(n);
    sam.solve();
    return ;
}
 
/*
*/