[ABC240E] Ranges on Tree 题解

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收起

[ABC240E] Ranges on Tree 题解

思路解析

由题意可知，只要一个点的所有儿子节点都被确定了，那么当前节点也就被确定了。也就是说，只要确定了所有叶子节点，也就能一层层地确定所有节点，而叶子节点没有儿子节点不受此条件的约束，同时我们希望 $\underset{i=1}{\overset{N}{max}}{R}_i$ 最小，所以我们把所有叶子节点的区间长度都置为 $1$ 即可。

code

#include<bits/stdc++.h>
using namespace std;
const int N = 2e5 + 10;
int n, fa[N], son[N], cnt = 0, l[N], r[N];
vector<int> g[N];
void dfs(int u, int f) {
	fa[u] = f;
	for(auto it : g[u]) if(it != f) son[u]++;
	if(son[u] == 0) l[u] = r[u] = ++cnt;
	else l[u] = 2e9, r[u] = 0;
	for(auto it : g[u]) {
		if(it != f) dfs(it, u), l[u] = min(l[u], l[it]), r[u] = max(r[u], r[it]);
	}
}
int main() {
	cin >> n;
	for(int i = 1, u, v; i < n; i++) {
		cin >> u >> v;
		g[u].push_back(v);
		g[v].push_back(u);
	}
	dfs(1, 0);
	for(int i = 1; i <= n; i++) {
		cout << l[i] << ' ' << r[i] << '\n';
	}
	return 0;
}