CF29D Ant on the Tree

题目

CF29D Ant on the Tree

算法标签: $dfs$, 树上差分

思路

由于点的数量很少$300$, 因此时间复杂度在$O(n^3)$之内都可以通过, 考虑$dfs$, 从叶子结点开始搜, 因为$dfs$记录顺序是逆序, 因此倒着搜就是正序, 时间复杂度$O(n^2)$

$dfs$代码

#include <iostream>
#include <algorithm>
#include <cstring>
#include <vector>

using namespace std;

const int N = 310;

vector<int> head[N], ans;

void add(int u, int v) {
	head[u].push_back(v);
}

bool dfs(int u, int fa, int t) {
	if (u == t) return true;

	for (int v : head[u]) {
		if (v != fa && dfs(v, u, t)) {
			ans.push_back(v);
			return true;
		}
	}
	return false;
}

int main() {
	ios::sync_with_stdio(false);
	cin.tie(0), cout.tie(0);

	int n;
	cin >> n;
	for (int i = 0; i < n - 1; ++i) {
		int u, v;
		cin >> u >> v;
		add(u, v), add(v, u);
	}

	int t = 1, s;
	while (cin >> s) dfs(s, 0, t), t = s;
	dfs(1, 0, t);

	ans.push_back(1);
	if (ans.size() != 2 * n - 1) {
		cout << -1 << "\n";
		return 0;
	}

	for (int x : ans) cout << x << " ";
	cout << "\n";
	return 0;
}

树上差分代码

在差分过程中, 计算每个节点应该经过的次数, 然后在统计的时候就能计算出每个节点应该被经过多少次, 从而判断是否是合法的

#include <iostream>
#include <algorithm>
#include <cstring>
#include <vector>

using namespace std;

const int N = 310, M = 10;

int n;
int fa[N][M], depth[N];
int diff[N], order[N], cnt;
vector<int> head[N];

void add(int u, int v) {
	head[u].push_back(v);
}

void dfs(int u, int pre, int dep) {
	fa[u][0] = pre, depth[u] = dep;
	int is_leaf = 1;
	for (int v : head[u]) {
		if (v == pre) continue;
		dfs(v, u, dep + 1);
		is_leaf = 0;
	}
	cnt += is_leaf;
}

int lca(int u, int v) {
	if (depth[u] < depth[v]) swap(u, v);
	for (int i = M - 1; i >= 0; --i) {
		if (depth[fa[u][i]] >= depth[v]) u = fa[u][i];
	}

	if (u == v) return u;

	for (int i = M - 1; i >= 0; --i) {
		if (fa[u][i] != fa[v][i]) {
			u = fa[u][i];
			v = fa[v][i];
		}
	}

	return fa[u][0];
}

void check(int u, int pre) {
	for (int v : head[u]) {
		if (v == pre) continue;
		check(v, u);
		diff[u] += diff[v];
	}
	if (u > 1 && diff[u] != 2) {
		cout << -1 << "\n";
		exit(0);
	}
}

void move(int u, int v) {
	int rev = 0;
	if (depth[u] < depth[v]) {
		swap(u, v);
		rev = 1;
	}

	vector<int> ans;
	while (u != v) {
		ans.push_back(rev ? u : fa[u][0]);
		u = fa[u][0];
	}

	if (rev) reverse(ans.begin(), ans.end());
	for (int u : ans) cout << u << " ";
}

int main() {
	ios::sync_with_stdio(false);
	cin.tie(0), cout.tie(0);

	cin >> n;
	for (int i = 0; i < n - 1; ++i) {
		int u, v;
		cin >> u >> v;
		add(u, v), add(v, u);
	}

	dfs(1, 0, 1);

	// 处理倍增数组
	for (int i = 1; i < M; ++i) {
		for (int u = 1; u <= n; ++u) {
			fa[u][i] = fa[fa[u][i - 1]][i - 1];
		}
	}

	// 起点终点都是1
	order[0] = order[cnt + 1] = 1;

	// 处理差分数组
	for (int i = 1; i <= cnt; ++i) {
		if (i <= cnt) cin >> order[i], diff[order[i]] += 2;
		int p = lca(order[i - 1], order[i]);
		diff[p] -= 2;
	}

	check(1, 0);

	cout << "1 ";
	for (int i = 1; i <= cnt + 1; ++i) {
		int p = lca(order[i - 1], order[i]);
		move(order[i - 1], p);
		move(p, order[i]);
	}
	cout << "\n";
	return 0;
}