Blog of RuSun

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

P2824 [HEOI2016/TJOI2016]排序

P2824 [HEOI2016/TJOI2016]排序

考虑二分答案 $k$ ,用 $0$ 表示 $\le k$ 的数,用 $1$ 表示 $ > k$ 的数,那么这样排序中只有 $0, 1$ ,可以用线段树区间修改、查询快速实现,如果位置 $q$ 不为 $0$ ,说明矛盾。发现有单调性,二分满足条件的最小的 $k$ 。

查看代码
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#include <cstdio>
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);
}
const int N = 1e5 + 10;
bool v[N];
int n, m, q, w[N];
struct Op { int op, l, r; } o[N];
struct Node { int l, r, tag, c; } tr[N << 2];
void pushup (int x) { tr[x].c = tr[x << 1].c + tr[x << 1 | 1].c; }
void cover (int x, bool k) { tr[x].c = (tr[x].r - tr[x].l + 1) * k, tr[x].tag = k; }
void pushdown (int x)
{
if (!~tr[x].tag) return;
cover(x << 1, tr[x].tag), cover(x << 1 | 1, tr[x].tag);
tr[x].tag = -1;
}
void build (int l = 1, int r = n, int x = 1)
{
tr[x].l = l, tr[x].r = r, tr[x].tag = -1;
if (l == r) return void(tr[x].c = v[l]);
int mid = l + r >> 1;
build(l, mid, x << 1), build(mid + 1, r, x << 1 | 1);
pushup(x);
}
void modify (int l, int r, bool k, int x = 1)
{
if (tr[x].l > r || l > tr[x].r) return;
if (tr[x].l >= l && tr[x].r <= r) return cover(x, k);
pushdown(x);
int mid = tr[x].l + tr[x].r >> 1;
modify(l, r, k, x << 1), modify(l, r, k, x << 1 | 1);
pushup(x);
}
int query (int l, int r, int x = 1)
{
if (tr[x].l > r || l > tr[x].r) return 0;
if (tr[x].l >= l && tr[x].r <= r) return tr[x].c;
pushdown(x);
int mid = tr[x].l + tr[x].r >> 1;
return query(l, r, x << 1) + query(l, r, x << 1 | 1);
}
bool check (int k)
{
for (int i = 1; i <= n; ++i) v[i] = w[i] > k;
build();
for (int i = 1; i <= m; ++i)
{
int c = query(o[i].l, o[i].r);
if (o[i].op == 0)
modify(o[i].l, o[i].r - c, 0), modify(o[i].r - c + 1, o[i].r, 1);
else if (o[i].op == 1)
modify(o[i].l, o[i].l + c - 1, 1), modify(o[i].l + c, o[i].r, 0);
}
return !query(q, q);
}
int main ()
{
read(n, m);
for (int i = 1; i <= n; ++i) read(w[i]);
for (int i = 1; i <= m; ++i)
read(o[i].op, o[i].l, o[i].r);
read(q);
int l = 1, r = n;
while (l < r)
{
int mid = l + r >> 1;
check(mid) ? r = mid : l = mid + 1;
}
write(l);
return 0;
}