「NOIP2016」天天爱跑步

传送门

Luogu

解题思路

树上差分+桶计数。

我们发现在一条路径上的点 \(i\) ,它可以观测到玩家的条件是：

• \(i \in (u \to LCA),dep_u=w_i+dep_i\)
• \(i \in (LCA \to v),dis(u,v)-dep_v=w_i-dep_i\)

所以我们就可以树上差分加桶计数实现，每次的答案就是对应桶的变化量。

细节注意事项

• 为了防止数组下标出现负数，可以统一加上一个数

参考代码

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <cctype>
#include <cmath>
#include <ctime>
#include <vector>
#define rg register
using namespace std;
template < typename T > inline void read(T& s) {
	s = 0; int f = 0; char c = getchar();
	while (!isdigit(c)) f |= (c == '-'), c = getchar();
	while (isdigit(c)) s = s * 10 + (c ^ 48), c = getchar();
	s = f ? -s : s;
}

const int _ = 300002;

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 n, m, w[_], ans[_], tong[2][_ * 3];
int dep[_], fa[21][_];
struct node { int f, p, v; } ;
vector < node > vec[_];

inline void dfs(int u, int f) {
	dep[u] = dep[f] + 1, fa[0][u] = f;
	for (rg int i = 1; i <= 19; ++i)
		fa[i][u] = fa[i - 1][fa[i - 1][u]];
	for (rg int i = head[u]; i; i = nxt[i]) {
		int v = ver[i]; if (v == f) continue;
		dfs(v, u);
	}
}

inline int LCA(int x, int y) {
	if (dep[x] < dep[y]) swap(x, y);
	for (rg int i = 19; ~i; --i)
		if (dep[fa[i][x]] >= dep[y]) x = fa[i][x];
	if (x == y) return x;
	for (rg int i = 19; ~i; --i)
		if (fa[i][x] != fa[i][y]) x = fa[i][x], y = fa[i][y];
	return fa[0][x];
}

inline void diff(int s, int t) {
	int lca = LCA(s, t), dis = dep[s] + dep[t] - 2 * dep[lca];
	vec[s].push_back((node) { 0, dep[s] + _, 1 });
	vec[lca].push_back((node) { 0, dep[s] + _, -1});
	vec[t].push_back((node) { 1, dis - dep[t] + _, 1 });
	vec[fa[0][lca]].push_back((node) { 1, dis - dep[t] + _, -1 });
}

inline void dfss(int u, int f) {
	int tmp = tong[0][w[u] + dep[u] + _] + tong[1][w[u] - dep[u] + _];
	for (rg int i = 0; i < (int) vec[u].size(); ++i)
		tong[vec[u][i].f][vec[u][i].p] += vec[u][i].v;
	for (rg int i = head[u]; i; i = nxt[i]) {
		int v = ver[i]; if (v == f) continue;
		dfss(v, u);
	}
	ans[u] = tong[0][w[u] + dep[u] + _] + tong[1][w[u] - dep[u] + _] - tmp;
}

int main() {
#ifndef ONLINE_JUDGE
	freopen("in.in", "r", stdin);
	freopen("out.out", "w", stdout);
#endif
	read(n), read(m);
	for (rg int u, v, i = 1; i < n; ++i)
		read(u), read(v), Add_edge(u, v), Add_edge(v, u);
	for (rg int i = 1; i <= n; ++i) read(w[i]);
	dfs(1, 0);
	for (rg int s, t, i = 1; i <= m; ++i) read(s), read(t), diff(s, t);
	dfss(1, 0);
	for (rg int i = 1; i <= n; ++i) printf("%d%c", ans[i], " \n"[i == n]);
	return 0;
}

完结撒花 \(qwq\)