Blog of RuSun

\begin {array}{c} \mathfrak {One Problem Is Difficult} \\\\ \mathfrak {Because You Don't Know} \\\\ \mathfrak {Why It Is Diffucult} \end {array}

SP220 PHRASES - Relevant Phrases of Annihilation

SP220 PHRASES - Relevant Phrases of Annihilation

SPOJ: PHRASES - Relevant Phrases of Annihilation

将所有串连起来,求 SA 。二分答案,注意到两个后缀的 LCP 是中间的所有 $height$ 的 $\min$ ,那么 $height < mid$ 的点会将整个序列划分为若干个部分,对于每个部分每个串求 $sa$ 的最值,即最大距离,如果对于每个串大于 $mid$ 即合法。复杂度 $O(n \log n)$ 。

查看代码
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#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
template <class Type>
void read (Type &x)
{
char c;
bool flag = false;
while ((c = getchar()) < '0' || c > '9')
c == '-' && (flag = true);
x = c - '0';
while ((c = getchar()) >= '0' && c <= '9')
x = (x << 1) + (x << 3) + c - '0';
flag && (x = ~x + 1);
}
template <class Type, class ...rest>
void read (Type &x, rest &...y) { read(x), read(y...); }
template <class Type>
void write (Type x)
{
x < 0 && (putchar('-'), x = ~x + 1);
x > 9 && (write(x / 10), 0);
putchar('0' + x % 10);
}
void chkmin (int &x, int k) { if (k < x) x = k; }
void chkmax (int &x, int k) { if (k > x) x = k; }
const int N = 5e5 + 10;
char str[N];
int n, m, id[N], sa[N], rnk[N], ht[N], x[N], y[N], c[N], mn[20], mx[20];
void bucsort ()
{
for (int i = 1; i <= m; ++i) c[i] = 0;
for (int i = 1; i <= n; ++i) ++c[x[i]];
for (int i = 2; i <= m; ++i) c[i] += c[i - 1];
for (int i = n; i; i--) sa[c[x[y[i]]]--] = y[i];
}
void init ()
{
for (int i = 1; i <= n; ++i)
x[i] = (str[i] - 'a' + 257) % 128, y[i] = i, sa[i] = 0;
bucsort();
for (int k = 1; k <= n; k <<= 1)
{
int cnt = 0;
for (int i = n - k + 1; i <= n; ++i) y[++cnt] = i;
for (int i = 1; i <= n; ++i)
if (sa[i] > k) y[++cnt] = sa[i] - k;
bucsort();
for (int i = 1; i <= n; ++i) y[i] = x[i];
x[sa[1]] = cnt = 1;
for (int i = 2; i <= n; ++i)
x[sa[i]] = cnt += (y[sa[i]] ^ y[sa[i - 1]] || y[sa[i] + k] ^ y[sa[i - 1] + k]);
if ((m = cnt) == n) break;
}
for (int i = 1; i <= n; ++i) rnk[sa[i]] = i;
for (int i = 1, k = 0; i <= n; ++i)
{
int j = sa[rnk[i] - 1];
if (k) --k;
while (i + k <= n && j + k <= n && str[i + k] == str[j + k]) ++k;
ht[rnk[i]] = k;
}
}
bool chk (int k, int mid)
{
bool flag = true;
for (int i = 1; i <= k && flag; ++i)
if (mx[i] - mn[i] < mid) flag = false;
for (int i = 1; i <= k; ++i)
mn[i] = n, mx[i] = 0;
return flag;
}
int main ()
{
int T; read(T);
while (T--)
{
n = 0;
int k; read(k);
for (int i = 1; i <= k; ++i)
{
scanf("%s", str + n + 1);
int t = strlen(str + n + 1);
for (int j = n + 1; j <= n + t; ++j) id[j] = i;
str[++(n += t)] = 'z' + i; id[n] = 0;
}
m = 26 + k;
init();
int l = 0, r = n >> 1;
while (l < r)
{
int mid = l + r + 1 >> 1;
bool flag = false;
for (int i = 2; i <= n && !flag; ++i)
{
if (ht[i] < mid) flag = chk(k, mid);
chkmin(mn[id[sa[i]]], sa[i]), chkmax(mx[id[sa[i]]], sa[i]);
}
chk(k, mid);
flag ? l = mid : r = mid - 1;
}
write(l), puts("");
}
return 0;
}