「JLOI2014」松鼠的新家

「JLOI2014」松鼠的新家

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两种做法：

• 树上差分 \(O(n)\)
• 树链剖分 \(O(nlogn)\)

树剖比较好写而且无脑，树上差分复杂度优秀一些但是会有点难调。

这里给出树剖写法：

唯一要讲的就是记得每次都把路径终点的贡献 \(-1\)

参考代码：

/*--------------------------------

  Code name: E.cpp

  Author: The Ace Bee

  This code is made by The Ace Bee

--------------------------------*/

#include <cstdio>

#define rg register

#define file(x)									\

	freopen(x".in", "r", stdin);				\

	freopen(x".out", "w", stdout);

const int $ = 500010;

inline void swap(int& a, int& b) { int t = a; a = b; b = t; }

inline int read() {

	int s = 0; bool f = false; char c = getchar();

	while (c < '0' || c > '9') f |= (c == '-'), c = getchar();

	while (c >= '0' && c <= '9') s = (s << 3) + (s << 1) + (c ^ 48), c = getchar();

	return f ? -s : s;

}

int n, a[$];

int tot, head[$], nxt[$ << 1], ver[$ << 1];

inline void Add_edge(int u, int v)

{ nxt[++tot] = head[u], head[u] = tot, ver[tot] = v; }

int siz[$], son[$], dep[$];

int top[$], rev[$], seg[$], father[$];

int sum[$ << 2], tag[$ << 2];

inline int lc(int rt) { return rt << 1; }

inline int rc(int rt) { return rt << 1 | 1; }

inline void pushup(int rt) {

	sum[rt] = sum[lc(rt)] + sum[rc(rt)];

}

inline void f(int rt, int l, int r, int v) {

	tag[rt] += v, sum[rt] += v * (r - l + 1);

}

inline void pushdown(int rt, int l, int r, int mid) {

	if (tag[rt]) f(lc(rt), l, mid, tag[rt]), f(rc(rt), mid + 1, r, tag[rt]), tag[rt] = 0;

}

inline void update(int rt, int l, int r, int x, int y, int v) {

	if (x <= l && r <= y) return f(rt, l, r, v);

	int mid = (l + r) >> 1;

	pushdown(rt, l, r, mid);

	if (x <= mid) update(lc(rt), l, mid, x, y, v);

	if (y > mid) update(rc(rt), mid + 1, r, x, y, v);

	pushup(rt);

}

inline int query(int rt, int l, int r, int id) {

	if (l == r) return sum[rt];

	int mid = (l + r) >> 1, res;

	pushdown(rt, l, r, mid);

	if (id <= mid) res = query(lc(rt), l, mid, id);

	else res = query(rc(rt), mid + 1, r, id);

	return res;

}

inline void dfs1(int u, int fa) {

	siz[u] = 1, father[u] = fa, dep[u] = dep[fa] + 1;

	for (rg int v, i = head[u]; i; i = nxt[i])

		if (!dep[v = ver[i]]) {

			dfs1(v, u), siz[u] += siz[v];

			if (siz[v] > siz[son[u]]) son[u] = v;

		}

}

inline void dfs2(int u, int topf) {

	top[rev[seg[u] = ++seg[0]] = u] = topf;

	if (!son[u]) return; dfs2(son[u], topf);

	for (rg int v, i = head[u]; i; i = nxt[i])

		if (!top[v = ver[i]]) dfs2(v, v);

}

inline void uptRange(int x, int y, int v) {

	int fx = top[x], fy = top[y];

	while (fx != fy) {

		if (dep[fx] < dep[fy]) swap(x, y), swap(fx, fy);

		update(1, 1, n, seg[fx], seg[x], v);

		x = father[fx], fx = top[x];

	}

	if (dep[x] > dep[y]) swap(x, y);

	update(1, 1, n, seg[x], seg[y], v);

}

int main() {

//	file("E");

	n = read();

	for (rg int i = 1; i <= n; ++i) a[i] = read();

	for (rg int u, v, i = 1; i <= n - 1; ++i)

		u = read(), v = read(), Add_edge(u, v), Add_edge(v, u);

	dfs1(1, 0), dfs2(1, 1);

	for (rg int i = 1; i < n; ++i)

		uptRange(a[i], a[i + 1], 1), uptRange(a[i + 1], a[i + 1], -1);

	for (rg int i = 1; i <= n; ++i)

		printf("%d\n", query(1, 1, n, seg[i]));

	return 0;

}