题目描述
Luogu 3975
BZOJ 3998
求一个串的第k小子串,不同位置的相同的子串算作一个或多个。
题解
看到子串就会想到后缀自动机,其实这个题只是一个后缀自动机的简单应用,只要运用一个搜索树的思想找第k个子串即可。
如果重复的子串算作一个的话,那就不用求endpos,直接设为1即可。否则就要求出endpos。
至于如何求某个节点x
“后面”的子串数tot
,就只用对tot[trans[x][i]]
求和即可。
Code
#include <cstdio>
#include <algorithm>
#include <cstring>
const int M = 1e6 + 10;
const int N = 5e5 + 10;
char s[N];
int trans[M][26], slink[M], maxlen[M], minlen[M], gr[M], ep[M], tot[M];
int len, n;
int tmp[N], rnk[M];
int t, k;
inline int new_state(int mx, int mn, int *tr, int sl) {
maxlen[n] = mx; minlen[n] = mn; slink[n] = sl;
for (int i = 0; i < 26; ++i) {
if (tr == NULL) trans[n][i]=-1;
else trans[n][i] = tr[i];
}
return n++;
}
inline void link(int x, int y) {
minlen[x] = maxlen[y]+1; slink[x] = y; deg[y]++;
}
inline int addc(int c, int u) {
int z = new_state(maxlen[u]+1, -1, NULL, -1);
gr[z] = 1;
while (u != -1 && trans[u][c] == -1) {
trans[u][c] = z;
u = slink[u];
}
if (u == -1) {
link(z, 0);
return z;
}
int x = trans[u][c];
if (maxlen[x] == maxlen[u]+1) {
link(z, x);
return z;
}
int y = new_state(maxlen[u]+1, minlen[x], trans[x], slink[x]);
link(x, y);
link(z, y);
while (u != -1 && trans[u][c] == x) {
trans[u][c] = y;
u = slink[u];
}
return z;
}
inline void build() {
int u = new_state(0, 0, NULL, -1);
for (int i = 0; i < len; ++i) u = addc(s[i]-'a', u);
for (int i = 0; i < n; ++i) tmp[maxlen[i]]++;
for (int i = 1; i <= len; ++i) tmp[i] += tmp[i-1];
for (int i = 0; i < n; ++i) rnk[tmp[maxlen[i]]--] = i;
for (int i = n; i; --i) {
int &j = rnk[i];
ep[j] += gr[j];
if (!t) ep[j] = 1;
else if (slink[j] != -1) ep[slink[j]] += ep[j];
}
ep[0] = 0;
for (int i = n; i; --i) {
int &j = rnk[i];
tot[j] = ep[j];
for (int c = 0; c < 26; ++c) if (trans[j][c] != -1)
tot[j] += tot[trans[j][c]];
}
}
int main() {
scanf("%s", s);
len = strlen(s);
scanf("%d%d", &t, &k);
build();
int x = 0;
if (k > tot[0]) {
puts("-1");
return 0;
}
while (1) {
for (int i = 0; i < 26; ++i) if (trans[x][i] != -1) {
int &y = trans[x][i];
if (k <= tot[y]) {
printf("%c", i+'a');
if (k <= ep[y]) return 0;
k -= ep[y];
x = y;
break;
} else k -= tot[y];
}
}
return 0;
}