[AcWing 3555] 二叉树

DFS + LCA

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点击查看代码

#include<bits/stdc++.h>

using namespace std;

typedef long long LL;

const int N = 1000 + 10;

int n, m;
int l[N], r[N], p[N];
int d[N];

void dfs(int u, int dist)
{
    d[u] = dist;
    if (l[u] != -1)
        dfs(l[u], dist + 1);
    if (r[u] != -1)
        dfs(r[u], dist + 1);
}

int lca(int a, int b)
{
    while (d[a] < d[b])
        b = p[b];
    while (d[a] > d[b])
        a = p[a];
    while (a != b) {
        a = p[a];
        b = p[b];
    }
    return a;
}

void solve()
{
    cin >> n >> m;
    for (int i = 1; i <= n; i ++) {
        int a, b;
        cin >> a >> b;
        l[i] = a, r[i] = b;
        if (a != -1)
            p[a] = i;
        if (b != -1)
            p[b] = i;
    }
    dfs(1, 0);
    while (m --) {
        int a, b;
        cin >> a >> b;
        int c = lca(a, b);
        cout << d[a] + d[b] - d[c] * 2 << endl;
    }
}

signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int T;
    cin >> T;
    while (T --) {
        solve();
    }

    // solve();

    return 0;
}

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1. 设两节点分别为 $a, b$，最近公共祖先为 $c$，根节点为 $r$，则
   $d_{ab} = d_{ar} + d_{br} - 2 \cdot d_{cr}$
2. 先用 $DFS$ 求出每个节点到根节点的距离，每次询问只需找到 $LCA$，代入上述公式即可
3. 寻找 $LCA$ 的过程：先让 $a$ 和 $b$ 处于同一层，然后让两节点同时移到其父节点，直到两节点相同