Blog of RuSun

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

P3759 [TJOI2017]不勤劳的图书管理员

P3759 [TJOI2017]不勤劳的图书管理员

动态逆序对问题,二维偏序,用分块套树状数组。合理控制块长,速度起飞。

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 = 5e4 + 10, B = 100, iB = 510, mod = 1e9 + 7;
int tot, id[N], st[N], ed[N];
int n, m, p[N], w[N], ans, tc[iB][N], ts[iB][N];
void adj (int &x) { x += x >> 31 & mod; }
void add (int x, int y, int op, int k)
{
for (int i = x; i <= tot; i += i & -i)
for (int j = y; j <= n; j += j & -j)
if (op) adj(tc[i][j] += 1 - mod), adj(ts[i][j] += k - mod);
else adj(--tc[i][j]), adj(ts[i][j] -= k);
}
void ask (int x, int y, int op, int &c, int &s)
{
for (int i = x; i; i -= i & -i)
for (int j = y; j; j -= j & -j)
if (op) adj(c += tc[i][j] - mod), adj(s += ts[i][j] - mod);
else adj(c -= tc[i][j]), adj(s -= ts[i][j]);
}
void query (int l, int r, int a, int b, int &c, int &s)
{
c = 0, s = 0;
if (l > r) return;
if (id[l] == id[r])
{
for (int i = l; i <= r; ++i)
if (p[i] >= a && p[i] <= b)
adj(c += 1 - mod), adj(s += w[i] - mod);
return;
}
for (int i = l; i <= ed[id[l]]; ++i)
if (p[i] >= a && p[i] <= b)
adj(c += 1 - mod), adj(s += w[i] - mod);
ask(id[r] - 1, b, 1, c, s), ask(id[l], b, 0, c, s), ask(id[r] - 1, a - 1, 0, c, s), ask(id[l], a - 1, 1, c, s);
for (int i = st[id[r]]; i <= r; ++i)
if (p[i] >= a && p[i] <= b)
adj(c += 1 - mod), adj(s += w[i] - mod);
}
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)
st[i] = ed[i - 1] + 1, ed[i] = ed[i - 1] + B;
ed[tot] = n;
for (int i = 1; i <= n; ++i)
{
read(p[i], w[i]);
int c, s;
query(1, i - 1, p[i] + 1, n, c, s);
ans = ((LL)c * w[i] + s + ans) % mod;
add(id[i], p[i], 1, w[i]);
}
for (int a, b; m; --m)
{
read(a, b);
if (a > b) swap(a, b);
int c, s;
if (p[a] > p[b]) adj(ans -= w[a] + w[b]);
query(a + 1, b - 1, 1, p[a] - 1, c, s); ans = (-(LL)c * w[a] - s + ans) % mod;
query(a + 1, b - 1, p[b] + 1, n, c, s); ans = (-(LL)c * w[b] - s + ans) % mod;
add(id[a], p[a], 0, w[a]), add(id[b], p[b], 0, w[b]);
swap(p[a], p[b]), swap(w[a], w[b]);
if (p[a] > p[b]) adj(ans += w[a] + w[b] - mod);
query(a + 1, b - 1, 1, p[a] - 1, c, s); ans = ((LL)c * w[a] + s + ans) % mod;
query(a + 1, b - 1, p[b] + 1, n, c, s); ans = ((LL)c * w[b] + s + ans) % mod;
add(id[a], p[a], 1, w[a]), add(id[b], p[b], 1, w[b]);
adj(ans), write(ans), puts("");
}
return 0;
}