CodeForces 1844G Tree Weights

洛谷传送门

CF 传送门

出题人脑洞真大……

设 $x_i$ 为 $i$ 到 $1$ 的距离。我们知道 $\forall i \in [1, n - 1], x_i + x_{i + 1} - 2 x_{\operatorname{lca}(i, i + 1)} = d_i$。发现每个方程有 $3$ 个未知数，很烦。能不能消掉一个？

发现若整个方程模 $2$，$x_{i + 1} \equiv d_i - x_i \pmod 2$。因此可以得知所有 $x_i$ 二进制下最低位。再递归处理接下来的位。于是我们发现 $x_i$ 若有解则一定是唯一确定的。

预处理出 $b_i = \operatorname{lca}(i, i + 1)$，时间复杂度 $O(n (\log n + \log V))$。

code

// Problem: G. Tree Weights
// Contest: Codeforces - Codeforces Round  884 (Div. 1 + Div. 2)
// URL: https://codeforces.com/contest/1844/problem/G
// Memory Limit: 256 MB
// Time Limit: 5000 ms
// 
// Powered by CP Editor (https://cpeditor.org)
 
#include <bits/stdc++.h>
#define pb emplace_back
#define fst first
#define scd second
#define mems(a, x) memset((a), (x), sizeof(a))
 
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef double db;
typedef long double ldb;
typedef pair<ll, ll> pii;
 
const int maxn = 100100;
 
ll n, a[maxn], b[maxn], c[maxn], d[maxn], ans[maxn];
vector<pii> G[maxn];
 
int fa[maxn], sz[maxn], son[maxn], dep[maxn], top[maxn];
bool vis[maxn];
 
int dfs(int u, int f, int d) {
	fa[u] = f;
	sz[u] = 1;
	dep[u] = d;
	int maxson = -1;
	for (pii p : G[u]) {
		int v = p.fst;
		if (v == f) {
			continue;
		}
		sz[u] += dfs(v, u, d + 1);
		if (sz[v] > maxson) {
			son[u] = v;
			maxson = sz[v];
		}
	}
	return sz[u];
}
 
void dfs2(int u, int tp) {
	top[u] = tp;
	vis[u] = 1;
	if (!son[u]) {
		return;
	}
	dfs2(son[u], tp);
	for (pii p : G[u]) {
		int v = p.fst;
		if (!vis[v]) {
			dfs2(v, v);
		}
	}
}
 
inline int qlca(int x, int y) {
	while (top[x] != top[y]) {
		if (dep[top[x]] < dep[top[y]]) {
			swap(x, y);
		}
		x = fa[top[x]];
	}
	if (dep[x] > dep[y]) {
		swap(x, y);
	}
	return x;
}
 
void dfs3(int u, int fa) {
	for (pii p : G[u]) {
		int v = p.fst, id = p.scd;
		if (v == fa) {
			continue;
		}
		ans[id] = a[v] - a[u];
		if (ans[id] <= 0) {
			puts("-1");
			exit(0);
		}
		dfs3(v, u);
	}
}
 
void solve() {
	scanf("%lld", &n);
	for (int i = 1, u, v; i < n; ++i) {
		scanf("%d%d", &u, &v);
		G[u].pb(v, i);
		G[v].pb(u, i);
	}
	for (int i = 1; i < n; ++i) {
		scanf("%lld", &d[i]);
	}
	dfs(1, -1, 1);
	dfs2(1, 1);
	for (int i = 1; i < n; ++i) {
		b[i] = qlca(i, i + 1);
	}
	for (int k = 0; k < 60; ++k) {
		for (int i = 1; i < n; ++i) {
			ll t = d[i] - a[i] - a[i + 1] + a[b[i]] * 2;
			if (t < 0 || t % (1LL << k)) {
				puts("-1");
				return;
			}
			t /= (1LL << k);
			t = (t % 2 + 2) % 2;
			c[i + 1] = ((t - c[i]) % 2 + 2) % 2;
		}
		for (int i = 2; i <= n; ++i) {
			if (c[i]) {
				a[i] |= (1LL << k);
			}
		}
	}
	dfs3(1, -1);
	for (int i = 1; i < n; ++i) {
		printf("%lld\n", ans[i]);
	}
}
 
int main() {
	int T = 1;
	// scanf("%d", &T);
	while (T--) {
		solve();
	}
	return 0;
}