Blog of RuSun

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

P3488 [POI2009]LYZ-Ice Skates

P3488 [POI2009]LYZ-Ice Skates

Hall定理:对于一个二分图,设左边有个 $n$ 点,右边有个 $m$ 点,则左边个点能完全匹配的充要条件是:对于 $1 \le i \le n$ ,左面任意 $i$ 个点,都至少有 $i$ 个右面的点与它相连。

记 $s _ i$ 为选择 $i$ 的数量,当前情况合法的条件:对于任意 $l, r$ ,

$$
\begin {aligned}
&\sum_{i = l} ^ r s _i \le k(r - l + 1 + d) \\
\Longrightarrow & \sum _ {i = l} ^ r s_i - k \le kd
\end {aligned}
$$
线段树维护 $s_i - k$ ,求最大子段和。

查看代码
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#include <cstdio>
#include <algorithm>
using namespace std;
typedef long long LL;
const int N = 5e5 + 10;
LL K, D;
int n, m;
struct Node
{
LL s, v, lv, rv;
int l, r;
} tr[N << 2];
void pushup(Node &x, Node l, Node r)
{
x.lv = max(l.lv, l.s + r.lv);
x.rv = max(r.rv, r.s + l.rv);
x.v = max(l.rv + r.lv, max(l.v, r.v));
x.s = l.s + r.s;
}
void build(int x, int l, int r)
{
tr[x].l = l;
tr[x].r = r;
if (l == r)
{
tr[x].s = tr[x].v = tr[x].lv = tr[x].rv = -K;
return;
}
int mid = l + r >> 1;
build(x << 1, l, mid);
build(x << 1 | 1, mid + 1, r);
pushup(tr[x], tr[x << 1], tr[x << 1 | 1]);
}
void modify(int x, int t, LL k)
{
if (tr[x].l == tr[x].r)
{
tr[x].s = tr[x].v = tr[x].lv = tr[x].rv += k;
return;
}
int mid = tr[x].l + tr[x].r >> 1;
if (t <= mid)
modify(x << 1, t, k);
else
modify(x << 1 | 1, t, k);
pushup(tr[x], tr[x << 1], tr[x << 1 | 1]);
}
Node query(int x, int l, int r)
{
if (tr[x].l >= l && tr[x].r <= r)
return tr[x];
int mid = tr[x].l + tr[x].r >> 1;
if (r <= mid)
return query(x << 1, l, r);
if (l > mid)
return query(x << 1 | 1, l, r);
Node res;
pushup(res, query(x << 1, l, r), query(x << 1 | 1, l, r));
return res;
}
int main()
{
scanf("%d%d%lld%lld", &n, &m, &K, &D);
build(1, 1, n);
for (int r, x; m; m--)
{
scanf("%d%d", &r, &x);
modify(1, r, x);
if (tr[1].v <= D * K)
puts("TAK");
else
puts("NIE");
}
return 0;
}