Codeforces round #628 C.Ehab and Path-etic MEXs

2024-05-08 05:08:54

Codeforces round #628 C.Ehab and Path-etic MEXs

You are given a tree consisting of n nodes. You want to write some labels on the tree’s edges such that the following conditions hold:

Every label is an integer between 0 and n−2 inclusive.
All the written labels are distinct.
The largest value among MEX(u,v) over all pairs of nodes (u,v) is as small as possible.
Here, MEX(u,v) denotes the smallest non-negative integer that isn’t written on any edge on the unique simple path from node u to node v.

Input

The first line contains the integer n (2≤n≤1e5) — the number of nodes in the tree.

Each of the next n−1 lines contains two space-separated integers u and v (1≤u,v≤n) that mean there’s an edge between nodes u and v. It’s guaranteed that the given graph is a tree.

Output

Output n−1 integers. The ith of them will be the number written on the ith edge (in the input order).

Examples

input

3
1 2
1 3

output

0
1

input

output

题目看似复杂其实很简单，我们考虑往边上填数字，有下面两种情况：
1.边的两个端点都不是叶子节点，即这条边为长链的一截，我们用贪心的策略不难发现，长链要往上面填大的数字，这样 MEX(u,v)MEX(u,v)MEX(u,v) 才能尽可能小
2.边上有一个结点为叶子节点，此时这条边构成一条长为1的短链，我们要从小往大填
这样就能构成一种最优解，AC代码如下：

#include<bits/stdc++.h>
using namespace std;
const int maxn=1e5+5;
vector<int>G[maxn];
struct node{int u,v;
}E[maxn];
int main()
{int n,u,v;scanf("%d",&n);for(int i=1;i<n;i++){scanf("%d%d",&u,&v);G[u].push_back(v);G[v].push_back(u);E[i].u=u,E[i].v=v;}int mx=n-2,mn=0;for(int i=1;i<n;i++){if(G[E[i].u].size()==1 || G[E[i].v].size()==1){printf("%d\n",mn);mn++;}else{printf("%d\n",mx);mx--;}}return 0;
}

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