题意
传送门 P6588『JROI-1』向量
题解
考虑相邻区间 A , B A,B A,B,尝试求解满足 i ∈ A , j ∈ B i\in A,j\in B i∈A,j∈B 对叉积的贡献
∑ i ∑ j a i ⃗ ⊕ a j ⃗ = ∑ i ∑ j ( x i y j − x j y i ) = ∑ i x i ∑ j y j − ∑ i y i ∑ j x j \sum_i\sum_j \vec{a_i}\oplus \vec{a_j}=\sum_i\sum_j(x_iy_j-x_jy_i)=\sum_i x_i\sum_j y_j-\sum_i y_i\sum_j x_j i∑j∑ai⊕aj=i∑j∑(xiyj−xjyi)=i∑xij∑yj−i∑yij∑xj 对点积的处理同理。线段树维护区间点积与叉积和的同时,维护区间 x , y x,y x,y 的和。总时间复杂度 O ( N log N ) O(N\log N) O(NlogN)。
#include <bits/stdc++.h>
using namespace std;
#define rep(i, l, r) for (int i = l, _ = r; i < _; ++i)
typedef long long ll;
const int maxn = 100005, sg_size = 1 << 18;
struct node
{
int sx, sy;
ll f, g;
} tree[sg_size];
int N, M, X[maxn], Y[maxn];
void up(node &chl, node &chr, node &p)
{
p.sx = chl.sx + chr.sx, p.sy = chl.sy + chr.sy;
p.f = chl.f + chr.f + (ll)chl.sx * chr.sx + (ll)chl.sy * chr.sy;
p.g = chl.g + chr.g + (ll)chl.sx * chr.sy - (ll)chl.sy * chr.sx;
}
void init(int k = 0, int l = 0, int r = N)
{
if (r - l == 1)
{
auto &u = tree[k];
u.sx = X[l], u.sy = Y[l], u.f = u.g = 0;
return;
}
int chl = (k << 1) + 1, chr = (k << 1) + 2, m = (l + r) >> 1;
init(chl, l, m), init(chr, m, r);
up(tree[chl], tree[chr], tree[k]);
}
void change(int a, int op, int x, int y, int k = 0, int l = 0, int r = N)
{
if (r - l == 1)
{
auto &u = tree[k];
if (op == 3)
u.sx *= x, u.sy *= x;
else
u.sx += x, u.sy += y;
return;
}
int chl = (k << 1) + 1, chr = (k << 1) + 2, m = (l + r) >> 1;
if (a < m)
change(a, op, x, y, chl, l, m);
if (m <= a)
change(a, op, x, y, chr, m, r);
up(tree[chl], tree[chr], tree[k]);
}
node ask(int a, int b, int k = 0, int l = 0, int r = N)
{
if (a <= l && r <= b)
return tree[k];
int chl = (k << 1) + 1, chr = (k << 1) + 2, m = (l + r) >> 1;
if (b <= m)
return ask(a, b, chl, l, m);
if (m <= a)
return ask(a, b, chr, m, r);
node res, rl = ask(a, b, chl, l, m), rr = ask(a, b, chr, m, r);
up(rl, rr, res);
return res;
}
int main()
{
ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
cin >> N >> M;
rep(i, 0, N) cin >> X[i] >> Y[i];
init();
while (M--)
{
int op, i, x, y, t, l, r;
cin >> op;
if (op == 1 || op == 2)
cin >> i >> x >> y, change(i - 1, op, op == 1 ? x : -x, op == 1 ? y : -y);
else if (op == 3)
cin >> i >> t, change(i - 1, op, t, 0);
else if (op == 4 || op == 5)
{
cin >> l >> r;
auto t = ask(l - 1, r);
cout << (op == 4 ? t.f : t.g) << '\n';
}
}
return 0;
}