[AcWing 3555] 二叉树
DFS + LCA
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#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;
}
- 设两节点分别为 \(a, b\),最近公共祖先为 \(c\),根节点为 \(r\),则
\(d_{ab} = d_{ar} + d_{br} - 2 \cdot d_{cr}\) - 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
- 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