[后缀自动机][单调队列优化DP] BZOJ 2806: [Ctsc2012]Cheat

Solution

对m个串建立后缀自动机。

对n个串单独考虑：

显然答案可以二分。

记pi表示这个串的i位置可以在SAM上匹配到的最长长度。

考虑DP：fi表示考虑前i个字符，最长熟悉字符串的长度。fi=max{fi−1,fj−j+i},L≤i−j≤pi分析一下有贡献的区间j∈[i−pi,i−L]。

pi有一个显然的性质pi≥pi+1−1。

考虑i和i+1：i+1−pi=i−(pi−1)≥i−pi所以单调队列维护递减的fi−i就可以O(n)DP啦。

#include <bits/stdc++.h>
#define show(x) cerr << #x << " = " << x << endl
using namespace std;
typedef long long ll;
typedef pair<int, int> Pairs;

const int N = 2202020;

inline char get(void) {
static char buf[100000], *S = buf, *T = buf;
if (S == T) {
T = (S = buf) + fread(buf, 1, 100000, stdin);
if (S == T) return EOF;
}
return *S++;
}
template<typename T>
inline void read(T &x) {
static char c; x = 0; int sgn = 0;
for (c = get(); c < '0' || c > '9'; c = get()) if (c == '-') sgn = 1;
for (; c >= '0' && c <= '9'; c = get()) x = x * 10 + c - '0';
if (sgn) x = -x;
}
inline void read(char *ch) {
char c; int len = 0;
for (c = get(); c < '0' || c > '1'; c = get());
for (; c >= '0' && c <= '1'; c = get()) ch[len++] = c;
ch[len] = 0;
}

int p
, f
;
int q
;
char s
;
namespace SAM {
int to
[4];
int mx
, par
;
int Tcnt, root, last;
inline void Init(void) { Tcnt = root = last = 1; }
inline int Extend(int c) {
int p = last, np = ++Tcnt;
mx[Tcnt] = mx[p] + 1;
for (; p && !to[p][c]; p = par[p])
to[p][c] = np;
if (p) {
int q = to[p][c];
if (mx[q] != mx[p] + 1) {
int nq = ++Tcnt;
mx[nq] = mx[p] + 1;
memcpy(to[nq], to[q], sizeof to[q]);
par[nq] = par[q];
par[q] = par[np] = nq;
for (; p && to[p][c] == q; p = par[p])
to[p][c] = nq;
} else {
par[np] = q;
}
} else {
par[np] = root;
}
return last = np;
}

int sta
, top;
inline void Debug(int u = 1) {
for (int i = 1; i <= top; i++) putchar(sta[i] + '0');
putchar('\n');
for (int i = 0; i < 3; i++)
if (to[u][i]) {
sta[++top] = i;
Debug(to[u][i]);
--top;
}
}
}
int n, m, len;

using namespace SAM;

inline bool Check(int L) {
int u = root;
for (int i = 1; i <= len; i++) {
int c = s[i] - '0';
if (to[u][c]) {
u = to[u][c]; p[i] = p[i - 1] + 1;
} else {
while (u && !to[u][c]) u = par[u];
if (u) {
p[i] = mx[u] + 1; u = to[u][c];
} else {
u = root; p[i] = 0;
}
}
}

for (int i = 0; i <= len; i++) f[i] = 0;
int h = 1, t = 0;
for (int i = 1; i <= len; i++) {
f[i] = f[i - 1];
if (i >= L) {
int res = f[i - L] - (i - L);
while (h <= t && f[q[t]] - q[t] < res) --t;
q[++t] = i - L;
}
while (h <= t && q[h] < i - p[i]) ++h;
if (h <= t) f[i] = max(f[i], f[q[h]] - q[h] + i);
}
return f[len] * 10 >= len * 9;
}

int main(void) {
freopen("1.in", "r", stdin);
read(n); read(m); Init();
for (int i = 1; i <= m; i++) {
read(s); len = strlen(s);
for (int i = 0; i < len; i++)
Extend(s[i] - '0');
Extend(2);
}
//  Debug();
for (int i = 1; i <= n; i++) {
read(s + 1); len = strlen(s + 1);
int L = 1, R = len, Mid, Ans;
while (L <= R) {
Mid = (L + R) >> 1;
if (Check(Mid)) L = (Ans = Mid) + 1;
else R = Mid - 1;
}
printf("%d\n", Ans);
}
return 0;
}