Blog of RuSun

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

P3203 [HNOI2010]弹飞绵羊

P3203 [HNOI2010]弹飞绵羊

注意坑点:题目中从 $0\ldots n - 1$ 。

对于 $x$ 的弹力为 $k$ ,从 $x$ 到 $x + k$ 连一条边,边界 $n + 1$ 时表示被弹飞。维护这些森林。查询时,将 $x$ 和 $n + 1$ 放在同一个 $splay$ 中,将 $x$ 放在根,$x$ 到 $n + 1$ 的节点数就是答案,即 $n + 1$ 节点左儿子的大小。

查看代码
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#include <cstdio>
#include <algorithm>
using namespace std;
const int N = 2e5 + 10;
int n, m, k[N];
struct Node
{
int sz;
int p, s[2], tag;
} tr[N];
void pushup(int x)
{
tr[x].sz = tr[tr[x].s[0]].sz + 1 + tr[tr[x].s[1]].sz;
}
void reverse(int x)
{
tr[x].tag ^= 1;
swap(tr[x].s[0], tr[x].s[1]);
}
void pushdown(int x)
{
if (tr[x].tag)
{
reverse(tr[x].s[0]);
reverse(tr[x].s[1]);
tr[x].tag = 0;
}
}
bool isroot(int x)
{
return tr[tr[x].p].s[0] != x && tr[tr[x].p].s[1] != x;
}
void rotate(int x)
{
int y = tr[x].p, z = tr[y].p;
int k = tr[y].s[1] == x;
if (!isroot(y))
tr[z].s[tr[z].s[1] == y] = x;
tr[x].p = z;
tr[y].s[k] = tr[x].s[k ^ 1];
tr[tr[x].s[k ^ 1]].p = y;
tr[x].s[k ^ 1] = y;
tr[y].p = x;
pushup(y);
pushup(x);
}
void update(int x)
{
if (!isroot(x))
update(tr[x].p);
pushdown(x);
}
void splay(int x)
{
update(x);
while (!isroot(x))
{
int y = tr[x].p, z = tr[y].p;
if (!isroot(y))
if ((y == tr[z].s[1]) ^ (x == tr[y].s[1]))
rotate(x);
else
rotate(y);
rotate(x);
}
}
void access(int x)
{
int tmp = x;
for (int y = 0; x; y = x, x = tr[x].p)
{
splay(x);
tr[x].s[1] = y;
pushup(x);
}
splay(tmp);
}
void mkroot(int x)
{
access(x);
reverse(x);
}
int findroot(int x)
{
access(x);
while (tr[x].s[0])
{
pushdown(x);
x = tr[x].s[0];
}
splay(x);
return x;
}
void split(int x, int y)
{
mkroot(x);
access(y);
}
void link(int x, int y)
{
mkroot(x);
if (findroot(y) != x)
tr[x].p = y;
}
void cut(int x, int y)
{
mkroot(x);
if (findroot(y) == x && tr[y].p == x && !tr[y].s[0])
{
tr[x].s[1] = tr[y].p = 0;
pushup(x);
}
}
int main()
{
scanf("%d", &n);
for (int i = 1; i <= n; i++)
{
scanf("%d", &k[i]);
link(i, min(i + k[i], n + 1));
}
scanf("%d", &m);
for (int op; m; m--)
{
scanf("%d", &op);
if (op == 1)
{
int x;
scanf("%d", &x);
x++;
split(x, n + 1);
printf("%d\n", tr[tr[n + 1].s[0]].sz);
}
else if (op == 2)
{
int x, t;
scanf("%d%d", &x, &t);
x++;
cut(x, min(x + k[x], n + 1));
k[x] = t;
link(x, min(x + k[x], n + 1));
}
}
return 0;
}