cf Educational Codeforces Round 78 E. Tests for problem D

原題：
E. Tests for problem D
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

We had a really tough time generating tests for problem D. In order to prepare strong tests, we had to solve the following problem.

Given an undirected labeled tree consisting of n

vertices, find a set of segments such that:

both endpoints of each segment are integers from 1 to 2n, and each integer from 1 to 2n should appear as an endpoint of exactly one segment; all segments are non-degenerate; for each pair (i,j)
such that i≠j, i∈[1,n] and j∈[1,n], the vertices i and j are connected with an edge if and only if the segments i and j intersect, but neither segment i is fully contained in segment j, nor segment j is fully contained in segment i

. 

Can you solve this problem too?
Input

The first line contains one integer n (1≤n≤5⋅10^5) — the number of vertices in the tree.

Then n−1 lines follow, each containing two integers xi and yi (1≤xi,yi≤n, xi≠yi) denoting the endpoints of the i-th edge.

It is guaranteed that the given graph is a tree.
Output

Print n pairs of integers, the i-th pair should contain two integers li and ri (1≤li<ri≤2n) — the endpoints of the i-th segment. All 2n integers you print should be unique.

It is guaranteed that the answer always exists.

Examples
Input

6
1 2
1 3
3 4
3 5
2 6

Output

9 12
7 10
3 11
1 5
2 4
6 8

Input

1

Output

1 2

中文：
C題的測試數據生成，給你一棵樹，現在讓你生成一堆線段，線段相交表示兩個節點相連。如果線段相互包含，則不算是兩個節點相連。

代碼：

    #include<iostream>
    #include<algorithm>
    #include<string>
    #include<vector>
    #include<string.h>
    #include<unordered_map>
    #include<map>
    #include<set>
     
     
    using namespace  std;
     
    typedef long long ll;
    typedef pair<int, int> pii;
    typedef pair<ll, ll> pll;
    const int maxn = 500010;
    typedef long long LL;
    vector<int> G[maxn];
     
    int n, cnt = 1;
    int ans[maxn][2];
     
     
    vector<pii> res;
    void dfs2(int cur, int pre, int len)
    {
    	int size;
    	if (cur == 1)
    		size = G[1].size();
    	else
    		size = G[cur].size() - 1;
    	cnt +=  size + 1;
    	//res.push_back(make_pair(len, cnt));
    	ans[cur][0] = len, ans[cur][1] = cnt;
    	int p = cnt;
    	for (int i = 0; i < G[cur].size(); i++)
    	{
    		if (G[cur][i] != pre)
    			dfs2(G[cur][i], cur, --p);
    	}
    }
    int main()
    {
    	ios::sync_with_stdio(false);
    	while (cin >> n)
    	{
    		int a, b;
    		res.clear();
    		for (int i = 1; i < n; i++)
    		{
    			cin >> a >> b;
    			G[a].push_back(b);
    			G[b].push_back(a);
    		}
    		dfs2(1, 0, 1);
    		for (int i = 1; i <= n;i++)
    		//for (int i = 0; i < res.size(); i++)
    			//cout << res[i].first << " " << res[i].second << endl;
    			cout << ans[i][0] << " " << ans[i][1] << endl;
    	}
    	return 0;
    }

思路：

比較簡單的構造題，首先找到當前節點有多少個子樹的個數，預留出子樹個數的長度加上2。同時，其子樹作爲線段的連接位置與長度，要再父節點的基礎上向右加寬即可。