「CC FAVNUM」Favourite Numbers-AC 自动机 + 数位 dp

给出一些数字串作为模式串,求 $[l, r]$ 中第 $k$ 个包含至少一个模式串的串。

链接

CC FAVNUM

题解

第一次知道 AC 自动机上还能数位 DP…

这个题还是很好想的,建出 AC 自动机,我们可以很轻松的求出 $[l, r]$ 中包含至少一个模式串的串的个数,为了方便我们可以先建出 Trie 图,并预处理出结束节点,然后数位 DP 时记录是否经过了这个节点。

然后题目要求第 $k$ 个串,我们二分或倍增就可以求出了。

代码

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/**
* Copyright (c) 2017, xehoth
* All rights reserved.
* 「CC FAVNUM」Favourite Numbers 26-09-2017
* AC 自动机 + 数位 dp
* @author xehoth
*/
#include <bits/stdc++.h>
namespace IO {
inline char read() {
static const int IN_LEN = 1000000;
static char buf[IN_LEN], *s, *t;
s == t ? t = (s = buf) + fread(buf, 1, IN_LEN, stdin) : 0;
return s == t ? -1 : *s++;
}
template <typename T>
inline void read(T &x) {
static char c;
static bool iosig;
for (iosig = false, c = read(); !isdigit(c); c = read()) {
if (c == -1) return;
c == '-' ? iosig = true : 0;
}
for (x = 0; isdigit(c); c = read()) x = x * 10 + (c ^ '0');
iosig ? x = -x : 0;
}
inline int read(char *buf) {
register int s = 0;
register char c;
while (c = read(), isspace(c) && c != -1)
;
if (c == -1) {
*buf = 0;
return -1;
}
do
buf[s++] = c;
while (c = read(), !isspace(c) && c != -1);
buf[s] = 0;
return s;
}
const int OUT_LEN = 10000000;
char obuf[OUT_LEN], *oh = obuf;
inline void print(char c) {
oh == obuf + OUT_LEN ? (fwrite(obuf, 1, OUT_LEN, stdout), oh = obuf) : 0;
*oh++ = c;
}
template <typename T>
inline void print(T x) {
static int buf[30], cnt;
if (x == 0) {
print('0');
} else {
x < 0 ? (print('-'), x = -x) : 0;
for (cnt = 0; x; x /= 10) buf[++cnt] = x % 10 | 48;
while (cnt) print((char)buf[cnt--]);
}
}
inline void print(const char *s) {
for (; *s; s++) print(*s);
}
inline void flush() { fwrite(obuf, 1, oh - obuf, stdout); }
struct InputOutputStream {
template <typename T>
inline InputOutputStream &operator>>(T &x) {
read(x);
return *this;
}
template <typename T>
inline InputOutputStream &operator<<(const T &x) {
print(x);
return *this;
}
~InputOutputStream() { flush(); }
} io;
}
namespace {
#define long long long
using IO::io;
const int MAXN = 2000;
const int MAX_SIGMA = 10;
struct Node {
Node *c[MAX_SIGMA], *fail;
int cnt;
inline void *operator new(size_t);
Node();
} pool[MAXN + 2], *null = pool, *cur = pool + 1;
inline void *Node::operator new(size_t) { return cur++; }
Node::Node() {
fail = null, cnt = 0;
for (register int i = 0; i < MAX_SIGMA; i++) c[i] = null;
}
struct AhoCrasickAutomation {
Node *root;
long f[20][MAXN + 2][2];
int digit[20];
AhoCrasickAutomation() {
root = new Node();
for (register int i = 0; i < MAX_SIGMA; i++) null->c[i] = root;
}
inline void build() {
static std::queue<Node *> q;
q.push(root);
for (Node *p, *u; !q.empty();) {
p = q.front(), q.pop();
for (register int i = 0; i < MAX_SIGMA; i++) {
if (p->c[i] != null) {
for (u = p->fail; u->c[i] == null;) u = u->fail;
p->c[i]->fail = u->c[i], q.push(p->c[i]);
p->c[i]->cnt |= u->c[i]->cnt;
} else {
p->c[i] = p->fail->c[i];
}
}
}
}
inline void insert(const char *s) {
Node *p = root;
for (; *s; s++) {
if (p->c[*s - '0'] == null) p->c[*s - '0'] = cur++;
p = p->c[*s - '0'];
}
p->cnt = 1;
}
inline long dfs(int pos, Node *p, int status, int limit) {
if (pos < 1) return status;
if (!limit && f[pos][p - pool][status] != -1)
return f[pos][p - pool][status];
register int end = limit ? digit[pos] : 9;
register long ret = 0;
for (register int i = 0; i <= end; i++)
ret += dfs(pos - 1, p->c[i], status || p->c[i]->cnt,
limit && i == end);
return !limit ? f[pos][p - pool][status] = ret : ret;
}
inline long solve(long x) {
register int len = 0;
for (; x; x /= 10) digit[++len] = x % 10;
return dfs(len, root, 0, 1);
}
inline void solve() {
register long l, r, k, n, step, w;
io >> l >> r >> k >> n;
static char buf[20];
for (register int i = 0; i < n; i++) io >> buf, insert(buf);
build();
memset(f, -1, sizeof(f));
k += solve(l - 1);
if (solve(r) < k) {
io << "no such number";
return;
}
step = 1, w = l - 1;
while (step < (r - l + 1)) step <<= 1;
while (step) {
if (solve(w + step) < k) w += step;
step >>= 1;
}
io << w + 1;
}
} task;
#undef long
}
int main() {
task.solve();
return 0;
}
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