【字尾自動機】 SPOJ SUBLEX

這題也是CLJ論文上的題目，做法CLJ論文講了。。。這裡不多說了。。。。

#include <iostream>
#include <queue> 
#include <stack> 
#include <map> 
#include <set> 
#include <bitset> 
#include <cstdio> 
#include <algorithm> 
#include <cstring> 
#include <climits>
#include <cstdlib>
#include <cmath>
#include <time.h>
#define maxn 200005
#define maxm 100005
#define eps 1e-7
#define mod 1000000007
#define INF 0x3f3f3f3f
#define PI (acos(-1.0))
#define lowbit(x) (x&(-x))
#define mp make_pair
#define ls o<<1
#define rs o<<1 | 1
#define lson o<<1, L, mid 
#define rson o<<1 | 1, mid+1, R
#define pii pair<int, int>
#pragma comment(linker, "/STACK:16777216")
typedef long long LL;
typedef unsigned long long ULL;
//typedef int LL;
using namespace std;
LL qpow(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base;base=base*base;b/=2;}return res;}
LL powmod(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base%mod;base=base*base%mod;b/=2;}return res;}
// head

struct node
{
	int len, cnt, vis;
	node *fa, *next[26];
}pool[maxn], *last, *root;

vector<int> ans;
char s[maxn];
int tot, m, k;

node* newnode(int len)
{
	node *p = &pool[tot++];
	p->len = len, p->cnt = p->vis = 0, p->fa = 0;
	memset(p->next, 0, sizeof p->next);
	return p;
}

void init()
{
	tot = 0;
	root = newnode(0);
	last = root;
}

void add(int c)
{
	node *p = last, *np = newnode(p->len + 1);
	last = np;
	for(; p && !p->next[c]; p = p->fa) p->next[c] = np;
	if(!p) np->fa = root;
	else {
		node *q = p->next[c];
		if(p->len + 1 == q->len) np->fa = q;
		else {
			node *nq = newnode(p->len + 1);
			*nq = *q;
			nq->len = p->len + 1;
			q->fa = np->fa = nq;
			for(; p && p->next[c] == q; p = p->fa) p->next[c] = nq;
		}
	}
}

void DFS(node *p)
{
	p->cnt = p->vis = 1;
	for(int i = 0; i < 26; i++) if(p->next[i]) {
		if(!p->next[i]->vis) DFS(p->next[i]);
		p->cnt += p->next[i]->cnt;
	}
}

void dfs(node *p)
{
	if(k <= 0) return;
	for(int i = 0; i < 26; i++) if(p->next[i]) {
		if(k > p->next[i]->cnt) k -= p->next[i]->cnt;
		else {
			ans.push_back('a' + i);
			k--;
			dfs(p->next[i]);
			break;
		}
	}
}

void read()
{
	scanf("%s%d", s, &m);
	for(int i = 0; s[i]; i++) add(s[i] - 'a');
}

void work()
{
	DFS(root);
	while(m--) {
		ans.clear();
		scanf("%d", &k);
		dfs(root);
		for(int i = 0; i < ans.size(); i++) printf("%c", ans[i]);
		printf("\n");
	}
}

int main()
{
	init();
	read();
	work();
	
	return 0;
}