Blog of RuSun

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

P3975 [TJOI2015]弦论

P3975 [TJOI2015]弦论

对于 $t = 0$ 这个问题是简单的。考虑 $t =1$ ,每个点的答案即 $endpos$ ,由于每个点的 $enpos$ 是子树的 $enpos$ 合并,那么可以求子树中叶子节点的个数,叶子节点即不能再被划分的点,即所有的原串前缀。那么问题转化为 $t = 0$ ,为了方便避免空串的影响,将 $f _ 1 = 0$ 。

查看代码
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#include <cstdio>
#include <string>
#include <vector>
#include <iostream>
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 << 3) + (x << 1) + c - '0';
if (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)
{
if (x < 0) putchar('-'), x = ~x + 1;
if (x > 9) write(x / 10);
putchar(x % 10 + '0');
}
typedef long long LL;
const int N = 1e6 + 10;
struct Node { int p, len, nxt[26]; } tr[N];
int hd = 1, tl, q[N];
int d[N], last = 1, cnt = 1;
LL h[N], f[N];
vector <int> g[N];
void insert(int c)
{
int p = last, np = ++cnt;
tr[np].len = tr[p].len + 1;
h[np] = 1;
for (; p && !tr[p].nxt[c]; p = tr[p].p) tr[p].nxt[c] = np;
last = np;
if (!p) return void(tr[np].p = 1);
int q = tr[p].nxt[c];
if (tr[q].len == tr[p].len + 1) return void(tr[np].p = q);
int nq = ++cnt;
tr[nq] = tr[q], tr[nq].len = tr[p].len + 1;
tr[q].p = tr[np].p = nq;
for (; p && tr[p].nxt[c] == q; p = tr[p].p) tr[p].nxt[c] = nq;
}
int main ()
{
string str;
cin >> str;
for (char c : str) insert(c - 'a');
for (int i = 1; i <= cnt; ++i)
for (int j = 0; j < 26; ++j)
if (tr[i].nxt[j])
g[i].push_back(j), ++d[tr[i].nxt[j]];
q[++tl] = 1;
while (hd <= tl)
{
int t = q[hd++];
f[t] = 1;
for (int i : g[t])
if (!--d[tr[t].nxt[i]])
q[++tl] = tr[t].nxt[i];
}
int T, k; read(T, k);
if (T) for (int i = hd; i; --i) h[tr[q[i]].p] += h[q[i]];
else for (int i = 1; i <= cnt; ++i) h[i] = 1;
h[1] = 0;
for (int i = 1; i <= cnt; ++i) f[i] = h[i];
for (int i = hd; i; --i)
for (int j : g[q[i]])
f[q[i]] += f[tr[q[i]].nxt[j]];
for (int p = 1; k <= f[p]; )
{
if ((k -= h[p]) <= 0) return 0;
for (int i : g[p])
if (f[tr[p].nxt[i]] >= k)
{
putchar(i + 'a');
p = tr[p].nxt[i];
break;
}
else k -= f[tr[p].nxt[i]];
}
puts("-1");
return 0;
}