割裂(Split Tree) 题解

思路分析

按题目要求使用对应数据结构与算法模拟/递推求解，注意数据范围边界与时间复杂度。

参考代码

#include <iostream>
#include <vector>
#include <queue>
#include <algorithm>
using namespace std;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int n, a;
    if (!(cin >> n >> a)) return 0;
    vector<int> head(n + 1, -1), to(2 * n), nxt(2 * n);
    int ec = 0;
    auto addEdge = [&](int u, int v) {
        to[ec] = v;
        nxt[ec] = head[u];
        head[u] = ec++;
    };
    for (int i = 1; i < n; ++i) {
        int u, v;
        cin >> u >> v;
        addEdge(u, v);
        addEdge(v, u);
    }
    vector<int> par(n + 1, 0), dep(n + 1, 0), sz(n + 1, 1), son(n + 1, 0), top(n + 1, 0), order;
    order.reserve(n);
    dep[1] = 1;
    order.push_back(1);
    for (int idx = 0; idx < (int)order.size(); ++idx) {
        int u = order[idx];
        for (int e = head[u]; e != -1; e = nxt[e]) {
            int v = to[e];
            if (!dep[v]) {
                dep[v] = dep[u] + 1;
                par[v] = u;
                order.push_back(v);
            }
        }
    }
    for (int i = n - 1; i >= 0; --i) {
        int u = order[i];
        int p = par[u];
        if (p) {
            sz[p] += sz[u];
            if (sz[u] > sz[son[p]]) son[p] = u;
        }
    }
    top[1] = 1;
    for (int i = 0; i < n; ++i) {
        int u = order[i];
        if (son[u]) top[son[u]] = top[u];
        for (int e = head[u]; e != -1; e = nxt[e]) {
            int v = to[e];
            if (v != par[u] && v != son[u]) top[v] = v;
        }
    }
    auto lca = [&](int u, int v) {
        while (top[u] != top[v]) {
            if (dep[top[u]] < dep[top[v]]) swap(u, v);
            u = par[top[u]];
        }
        return dep[u] < dep[v] ? u : v;
    };
    vector<int> diff(n + 1, 0);
    for (int i = 0; i < a; ++i) {
        int u, v;
        cin >> u >> v;
        int l = lca(u, v);
        diff[u]++;
        diff[v]++;
        diff[l]--;
        if (par[l]) diff[par[l]]--;
    }
    int bu, bv;
    cin >> bu >> bv;
    for (int i = n - 1; i >= 0; --i) {
        int u = order[i];
        if (par[u]) diff[par[u]] += diff[u];
    }
    int bl = lca(bu, bv);
    int ans = 0;
    for (int cur = bu; cur != bl; cur = par[cur]) {
        if (diff[cur] == 0) ++ans;
    }
    for (int cur = bv; cur != bl; cur = par[cur]) {
        if (diff[cur] == 0) ++ans;
    }
    if (diff[bl] == 0) ++ans;
    cout << ans << "\n";
    return 0;
}