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

CF1093E Intersection of Permutations

CF1093E Intersection of Permutations

将问题转化为 $[l, r]$ 有多少个数在 $[a, b]$ 之间。二位偏序,这里用分块套树状数组。

查看代码
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#include <cstdio>
#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);
}
typedef long long LL;
const int N = 2e5 + 10, B = 450;
int n, m, w[N], p[N], tot, id[N], L[N], R[N], tr[B][N];
void add (int x, int y, int k)
{
for (int i = x; i <= tot; i += i & -i)
for (int j = y; j <= n; j += j & -j)
tr[i][j] += k;
}
int query (int x, int y)
{
int res = 0;
for (int i = x; i; i -= i & -i)
for (int j = y; j; j -= j & -j)
res += tr[i][j];
return res;
}
int main ()
{
read(n, m);
for (int i = 1; i <= n; ++i)
id[i] = (i - 1) / B + 1;
tot = id[n];
for (int i = 1; i <= tot; ++i)
L[i] = R[i - 1] + 1, R[i] = R[i - 1] + B;
R[tot] = n;
for (int i = 1, a; i <= n; ++i)
read(a), p[a] = i;
for (int i = 1; i <= n; ++i)
read(w[i]), add(id[i], w[i] = p[w[i]], 1);
for (int op; m; --m)
{
read(op);
if (op == 1)
{
int a, b, l, r; read(a, b, l, r);
if (id[l] == id[r])
{
int res = 0;
for (int i = l; i <= r; ++i)
res += w[i] >= a && w[i] <= b;
write(res), puts("");
}
else
{
int res = 0;
for (int i = l; i <= R[id[l]]; ++i)
res += w[i] >= a && w[i] <= b;
res += query(id[r] - 1, b) - query(id[r] - 1, a - 1) - query(id[l], b) + query(id[l], a - 1);
for (int i = L[id[r]]; i <= r; ++i)
res += w[i] >= a && w[i] <= b;
write(res), puts("");
}
}
else if (op == 2)
{
int a, b; read(a, b);
add(id[a], w[a], -1), add(id[b], w[b], -1);
swap(w[a], w[b]);
add(id[a], w[a], 1), add(id[b], w[b], 1);
}
}
return 0;
}