C. Ehab and Path-etic MEXs
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
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≤105) — 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
inputCopy
3
1 2
1 3
outputCopy
0
1
inputCopy
6
1 2
1 3
2 4
2 5
5 6
outputCopy
0
3
2
4
1
Note
The tree from the second sample:
思路：考慮兩種極端情況把
第一種 如果樹退化成鏈，那麼其實不用管，不論怎麼輸出mex最大值都是一樣的 都是n-1
因爲鏈的兩個端點之間 一定是從0-n-2所有數 所以mex一定是n-1
第二種 就是考慮只有一個父節點，第二層都是子節點，也就是說父節點出度爲n-1，其他點入度爲1 這樣子的mex最大值最小是2，因爲任意兩點之間最多經過2個點，推廣到不是第一種極端情況的所有情況，其實mex的最大值最小一定是2，只要保證0 1 2三個點不在同一條鏈上即可
所以問題轉化成怎麼讓0 1 2不形成一個鏈，考慮度，如果一個點k，有三個邊與他相連，我們把0 1 2放再他的三個相連邊即可
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define me(a,x) memset(a,x,sizeof a)
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define all(x) (x).begin(), (x).end()
#define pb(a) push_back(a)
#define paii pair<int,int>
#define pali pair<ll,int>
#define pail pair<int,ll>
#define pall pair<ll,ll>
#define x first
#define y second
int d[100005];
paii v[100005];
bool vis[100005];
int main()
{
int n;cin>>n;
rep(i,1,n){
cin>>v[i].x>>v[i].y;
d[v[i].y]++;
d[v[i].x]++;
}
int k=0;
repd(i,1,n){
if(d[i]>=3){
k=i;
break;
}
}
if(k==0){
rep(i,1,n) cout<<i-1<<endl;
}
else
{
int num=0,q=3;
rep(i,1,n){
if(num<3 && (v[i].x==k || v[i].y==k)){
cout<<num<<endl;
num++;
}
else cout<<q++<<endl;
}
}
return 0;
}