KDTree(Bzoj2648: SJY摆棋子)
题面
KDTree
大概就是一个分割\(k\)维空间的数据结构，二叉树
建立：每层选取一维为关键字，把中间的点拿出来，递归左右，有个\(STL\)函数nth_element可以用一下
维护：维护当前这个点的子树的每一维的最大值和最小值，相当于维护了个高维矩形
查询：直接遍历一棵树是\(O(n)\)的，利用一些独特的性质可以剪枝，因题而异
奇技淫巧：
- 把坐标绕原点转\(\alpha\)度
- 定期重构或者像替罪羊树一样，利用平衡因子判断是否需重构
- 每次查询到某一层优先选取答案可能最优的一个儿子先递归
- 等等
Sol
\(KDTree\)模板题
利用替罪羊树的思想重构
# include <bits/stdc++.h>
# define IL inline
# define RG register
# define Fill(a, b) memset(a, b, sizeof(a))
using namespace std;
typedef long long ll;
IL int Input(){
RG int x = 0, z = 1; RG char c = getchar();
for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;
for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
return x * z;
}
const int maxn(1e6 + 5);
const int inf(2e9);
const double alpha(0.75);
int n, op, rt, q, ans, p[maxn], tot, num, cnt;
struct Point{
int d[2];
IL int operator <(RG Point a) const{
return d[op] < a.d[op];
}
} a[maxn];
struct KDTree{
int ch[2], d[2], size;
int mn[2], mx[2];
} tr[maxn];
IL void Chkmax(RG int &x, RG int y){
if(y > x) x = y;
}
IL void Chkmin(RG int &x, RG int y){
if(y < x) x = y;
}
IL int NewNode(){
return tot ? p[tot--] : ++cnt;
}
IL void Update(RG int x){
RG int ls = tr[x].ch[0], rs = tr[x].ch[1];
tr[x].mx[0] = tr[x].mn[0] = tr[x].d[0];
tr[x].mx[1] = tr[x].mn[1] = tr[x].d[1];
tr[x].size = tr[ls].size + tr[rs].size + 1;
if(ls){
Chkmin(tr[x].mn[0], tr[ls].mn[0]), Chkmin(tr[x].mn[1], tr[ls].mn[1]);
Chkmax(tr[x].mx[0], tr[ls].mx[0]), Chkmax(tr[x].mx[1], tr[ls].mx[1]);
}
if(rs){
Chkmin(tr[x].mn[0], tr[rs].mn[0]), Chkmin(tr[x].mn[1], tr[rs].mn[1]);
Chkmax(tr[x].mx[0], tr[rs].mx[0]), Chkmax(tr[x].mx[1], tr[rs].mx[1]);
}
}
IL int Build(RG int l, RG int r, RG int nop){
op = nop;
RG int x = (l + r) >> 1, nw = NewNode();
nth_element(a + l, a + x, a + r + 1);
tr[nw].d[0] = a[x].d[0], tr[nw].d[1] = a[x].d[1];
if(l < x) tr[nw].ch[0] = Build(l, x - 1, nop ^ 1);
if(x < r) tr[nw].ch[1] = Build(x + 1, r, nop ^ 1);
Update(nw);
return nw;
}
IL int Check(RG int x){
return alpha * tr[x].size < tr[tr[x].ch[0]].size || alpha * tr[x].size < tr[tr[x].ch[1]].size;
}
IL void ReCycle(RG int x){
if(!x) return;
ReCycle(tr[x].ch[0]), tr[x].ch[0] = 0;
p[++tot] = x, a[++num].d[0] = tr[x].d[0], a[num].d[1] = tr[x].d[1];
ReCycle(tr[x].ch[1]), tr[x].ch[1] = 0;
}
IL int ReBuild(RG int x, RG int nop){
num = 0, ReCycle(x);
return Build(1, num, nop);
}
IL void Insert(RG int &x, RG int nop, RG Point np){
if(!x){
x = NewNode();
tr[x].d[0] = np.d[0], tr[x].d[1] = np.d[1];
tr[x].mx[0] = tr[x].mn[0] = tr[x].d[0];
tr[x].mx[1] = tr[x].mn[1] = tr[x].d[1];
return;
}
if(np.d[nop] < tr[x].d[nop]) Insert(tr[x].ch[0], nop ^ 1, np);
else Insert(tr[x].ch[1], nop ^ 1, np);
Update(x);
if(Check(x)) x = ReBuild(x, nop);
}
IL int Calc(RG int x, RG Point np){
RG int d = max(np.d[0] - tr[x].mx[0], 0) + max(tr[x].mn[0] - np.d[0], 0);
return d + max(tr[x].mn[1] - np.d[1], 0) + max(np.d[1] - tr[x].mx[1], 0);
}
IL void Query(RG int x, RG Point np){
RG int d = abs(tr[x].d[0] - np.d[0]) + abs(tr[x].d[1] - np.d[1]), d1 = inf, d2 = inf;
Chkmin(ans, d);
if(tr[x].ch[0]) d1 = Calc(tr[x].ch[0], np);
if(tr[x].ch[1]) d2 = Calc(tr[x].ch[1], np);
RG int c = 0;
if(d1 > d2) swap(d1, d2), c ^= 1;
if(d1 < ans) Query(tr[x].ch[c], np);
if(d2 < ans) Query(tr[x].ch[!c], np);
}
int main(){
n = Input(), q = Input();
for(RG int i = 1; i <= n; ++i) a[i].d[0] = Input(), a[i].d[1] = Input();
rt = Build(1, n, 0);
for(RG int i = 1; i <= q; ++i){
RG int t = Input(), x = Input(), y = Input();
if(t == 1) Insert(rt, 0, (Point){x, y});
else{
ans = inf;
Query(rt, (Point){x, y});
printf("%d\n", ans);
}
}
return 0;
}
