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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1order[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;
}