「BZOJ 4154」Generating Synergy-k-d 树

给定一棵以 $1$ 为根的有根树,初始所有节点颜色为 $1$,每次将距离节点 $a$ 不超过 $l$ 的 $a$ 的子节点染成 $c$,或询问点 $a$ 的颜色。

链接

BZOJ 4154

题解

子树修改,首先肯定会想到 dfs 序,但此题还有距离限制,直接 dfs 序肯定是不行的,据说可以点分???

这个题有一个巧妙的做法,我们把 dfs 序看做横坐标,深度看做纵坐标,那么每次修改就是将一个矩形内的点染成一种颜色,即将矩形 $(dfn[a], dep[a]), (dfn[a] + sz[a] - 1, dep[a] + l)$ 染成颜色 $c$,所以我们用 K-D 树维护一个染色标记就可以实现。

一些细节:我们可以记录 K-D 树的父亲节点及原树中节点对应 K-D 树中的节点,这样查询的时候我们只需要直接找到对应 K-D 树中的节点,然后沿 fa 跳回根下传标记即可。

另外这题数据水到了按小的方差划分也能过……

代码

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/**
* Copyright (c) 2017, xehoth
* All rights reserved.
* 「BZOJ 4154」Generating Synergy 02-10-2017
* K-D 树
* @author xehoth
*/
#include <bits/stdc++.h>
namespace IO {
inline char read() {
static const int IN_LEN = 100000;
static char buf[IN_LEN], *s, *t;
s == t ? t = (s = buf) + fread(buf, 1, IN_LEN, stdin) : 0;
return s == t ? -1 : *s++;
}
template <typename T>
inline void read(T &x) {
static char c;
static bool iosig;
for (c = read(), iosig = false; !isdigit(c); c = read()) {
if (c == -1) return;
c == '-' ? iosig = true : 0;
}
for (x = 0; isdigit(c); c = read()) x = x * 10 + (c ^ '0');
iosig ? x = -x : 0;
}
inline int read(char *buf) {
register int s = 0;
register char c;
while (c = read(), isspace(c) && c != -1)
;
if (c == -1) {
*buf = 0;
return -1;
}
do
buf[s++] = c;
while (c = read(), !isspace(c) && c != -1);
buf[s] = 0;
return s;
}
inline void read(char &x) {
while (x = read(), isspace(x) && x != -1)
;
}
const int OUT_LEN = 100000;
char obuf[OUT_LEN], *oh = obuf;
inline void print(char c) {
oh == obuf + OUT_LEN ? (fwrite(obuf, 1, oh - obuf, stdout), oh = obuf) : 0;
*oh++ = c;
}
template <typename T>
inline void print(T x) {
static int buf[30], cnt;
if (x == 0) {
print('0');
} else {
x < 0 ? (print('-'), x = -x) : 0;
for (cnt = 0; x; x /= 10) buf[++cnt] = x % 10 | 48;
while (cnt) print((char)buf[cnt--]);
}
}
inline void flush() { fwrite(obuf, 1, oh - obuf, stdout); }
struct InputOutputStream {
template <typename T>
inline InputOutputStream &operator<<(const T &x) {
print(x);
return *this;
}
template <typename T>
inline InputOutputStream &operator>>(T &x) {
read(x);
return *this;
}
~InputOutputStream() { flush(); }
} io;
}
namespace {
#define long long long
const int MOD = 1e9 + 7;
const int MAXN = 100010;
struct IdPoint {
int x, y, id;
IdPoint(int x = 0, int y = 0, int id = 0) : x(x), y(y), id(id) {}
} p[MAXN + 1];
struct Point {
int x, y;
Point(int x = 0, int y = 0) : x(x), y(y) {}
Point(const IdPoint &p) : x(p.x), y(p.y) {}
};
struct Node {
Node *c[2], *fa;
Point p, min, max;
int cov, color;
Node();
Node(const IdPoint &);
inline void maintain() {
min.x = std::min(std::min(c[0]->min.x, c[1]->min.x), min.x);
min.y = std::min(std::min(c[0]->min.y, c[1]->min.y), min.y);
max.x = std::max(std::max(c[0]->max.x, c[1]->max.x), max.x);
max.y = std::max(std::max(c[0]->max.y, c[1]->max.y), max.y);
}
inline void cover(int);
inline void pushDown() {
if (cov) c[0]->cover(cov), c[1]->cover(cov), cov = 0;
}
inline void *operator new(size_t);
} pool[MAXN + 1], *cur = pool + 1, *null = pool, *id[MAXN + 1];
Node::Node() : cov(0), color(0) {
c[0] = c[1] = fa = null;
min.x = min.y = INT_MAX, max.x = max.y = INT_MIN;
}
Node::Node(const IdPoint &p)
: p(p), min(p), max(p), cov(0), color(1), fa(null) {
c[0] = c[1] = null;
}
inline void *Node::operator new(size_t) { return cur++; }
inline void Node::cover(int cov) {
if (this == null) return;
this->color = this->cov = cov;
}
bool flag;
inline bool cmp(const IdPoint &p1, const IdPoint &p2) {
return flag ? (p1.y < p2.y || (p1.y == p2.y && p1.x < p2.x))
: (p1.x < p2.x || (p1.x == p2.x && p1.y < p2.y));
}
template <typename T>
inline T square(const T &x) {
return x * x;
}
inline bool getSplit(int l, int r) {
register double vx = 0, vy = 0, ax = 0, ay = 0;
for (register int i = l; i <= r; i++) ax += p[i].x, ay += p[i].y;
ax /= r - l + 1, ay /= r - l + 1;
for (register int i = l; i <= r; i++)
vx += square(p[i].x - ax), vy += square(p[i].y - ay);
return vx < vy;
}
inline Node *build(int l, int r) {
if (l > r) return null;
register int mid = l + r >> 1;
flag = getSplit(l, r), std::nth_element(p + l, p + mid, p + r + 1, cmp);
register Node *o = new Node(p[mid]);
id[p[mid].id] = o;
o->c[0] = build(l, mid - 1), o->c[1] = build(mid + 1, r);
if (o->c[0] != null) o->c[0]->fa = o;
if (o->c[1] != null) o->c[1]->fa = o;
return o->maintain(), o;
}
std::vector<int> edge[MAXN + 1];
typedef std::vector<int>::iterator Iterator;
inline void addEdge(const int u, const int v) {
edge[u].push_back(v), edge[v].push_back(u);
}
int dfn[MAXN + 1], sz[MAXN + 1], dep[MAXN + 1], fa[MAXN + 1], idx;
bool vis[MAXN + 1];
inline void dfs(const int u) {
vis[u] = true, dep[u] = dep[fa[u]] + 1, sz[u] = 1, dfn[u] = ++idx;
for (Iterator v = edge[u].begin(); v != edge[u].end(); v++)
if (!vis[*v]) fa[*v] = u, dfs(*v), sz[u] += sz[*v];
}
inline void init(const int n) {
for (register int i = 0; i <= n; i++) edge[i].clear();
idx = 0;
memset(vis, 0, sizeof(bool) * (n + 1)), fa[1] = 0;
cur = pool + 1;
}
inline void modify(Node *p, const Point &l, const Point &r, int color) {
if (p == null || p->min.x > r.x || p->max.x < l.x || p->min.y > r.y ||
p->max.y < l.y)
return;
p->pushDown();
if (p->max.x <= r.x && p->min.x >= l.x && p->max.y <= r.y &&
p->min.y >= l.y) {
p->cov = p->color = color;
return;
}
if (p->p.x <= r.x && p->p.x >= l.x && p->p.y <= r.y && p->p.y >= l.y)
p->color = color;
modify(p->c[0], l, r, color), modify(p->c[1], l, r, color);
}
inline int query(Node *p) {
static Node *st[MAXN + 1];
register int top = 0;
for (; p != null; p = p->fa) st[++top] = p;
while (top) st[top]->pushDown(), top--;
return st[1]->color;
}
using IO::io;
inline void solve() {
register int T;
io >> T;
while (T--) {
register int n, c, q;
io >> n >> c >> q;
init(n);
for (register int i = 2, v; i <= n; i++) io >> v, addEdge(i, v);
dfs(1);
for (register int i = 1; i <= n; i++)
p[i].x = dfn[i], p[i].y = dep[i], p[i].id = i;
register Node *root = build(1, n);
register int ans = 0;
for (register int x, y, cmd, i = 1; i <= q; i++) {
io >> x >> y >> cmd;
switch (cmd) {
case 0:
ans = (ans + (long)query(id[x]) * i) % MOD;
break;
default:
modify(root, Point(dfn[x], dep[x]),
Point(dfn[x] + sz[x] - 1, dep[x] + y), cmd);
break;
}
}
io << ans << '\n';
}
}
#undef long
}
int main() {
#ifdef DBG
freopen("sample/1.in", "r", stdin);
#endif
solve();
return 0;
}
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