SPOJ：SUBLEX

题面

第一行给定主串((len<=90000))
第二行给定询问个数(T<=500)
随后给出(T)行(T)个询问，每次询问排名第(k)小的串，范围在(int)内
相同的子串算一个

Sol

(sam)中每条从起点出发的路径都对应一个子串
设(f[i])表示从(i)出发的路径的条数
(f[i]=1+sum_{jin trans[i]}f[j])
最后贪心一遍就好了

# include <bits/stdc++.h>
# define IL inline
# define RG register
# define Fill(a, b) memset(a, b, sizeof(a))
# define File(a) freopen(a".in", "r", stdin), freopen(a".out", "w", stdout)
using namespace std;
typedef long long ll;

IL int Input(){
	RG int x = 0, z = 1; RG char c = getchar();
	for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;
	for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
	return x * z;
}

const int maxn(2e5 + 5);

int trans[26][maxn], fa[maxn], len[maxn], tot = 1, last = 1;
int n, f[maxn], t[maxn], id[maxn];
char s[maxn];

IL void Extend(RG int c){
	RG int p = last, np = ++tot; last = tot;
	len[np] = len[p] + 1;
	while(p && !trans[c][p]) trans[c][p] = np, p = fa[p];
	if(!p) fa[np] = 1;
	else{
		RG int q = trans[c][p];
		if(len[q] == len[p] + 1) fa[np] = q;
		else{
			RG int nq = ++tot;
			len[nq] = len[p] + 1, fa[nq] = fa[q];
			for(RG int i = 0; i < 26; ++i) trans[i][nq] = trans[i][q];
			fa[q] = fa[np] = nq;
			while(p && trans[c][p] == q) trans[c][p] = nq, p = fa[p];
		}
	}
}

IL void Calc(RG int k){
	RG int nw = 1;
	while(k){
		for(RG int i = 0; i < 26; ++i){
			RG int p = trans[i][nw];
			if(p){
				if(k <= f[p]){
					--k, putchar(i + 'a'), nw = p;
					break;
				}
				else k -= f[p];
			}
		}
	}
	puts("");
}

int main(RG int argc, RG char* argv[]){
	scanf(" %s", s), n = strlen(s);
	for(RG int i = 0; i < n; ++i) Extend(s[i] - 'a');
	for(RG int i = 1; i <= tot; ++i) ++t[len[i]], f[i] = 1;
	for(RG int i = 1; i <= tot; ++i) t[i] += t[i - 1];
	for(RG int i = 1; i <= tot; ++i) id[t[len[i]]--] = i;
	for(RG int i = tot; i; --i)
		for(RG int j = 0; j < 26; ++j) f[id[i]] += f[trans[j][id[i]]];
	for(RG int T = Input(); T; --T) Calc(Input());
	return 0;
}