题意：给你一棵树，n点n-1条边，现在这些边都不存在，你要修起来，你可以选择一条边修理，如果这个边的两边的城市都没有在修理其他边的话。每次修理需要一天的时间。问你最少多少天可以修完，并且把方案输出。
思路：dfs一波
#include<bits/stdc++.h>
using namespace std;
const int maxn = 2e5+10;
vector<pair<int,int> >e[maxn];
vector<int>ans[maxn];
int tot = 0;
void dfs(int x,int fa,int t)
{
int now = 0;
for (int i = 0;i<e[x].size();i++)
{
int v = e[x][i].first;
if (v==fa)
continue;
now++;
if (now == t)
now++;
tot = max(tot,now);
ans[now].push_back(e[x][i].second);
dfs(v,x,now);
}
}
int main()
{
int n;
scanf("%d",&n);
for (int i = 1;i<n;i++)
{
int u,v;
scanf("%d%d",&u,&v);
e[u].push_back(make_pair(v,i));
e[v].push_back(make_pair(u,i));
}
dfs(1,-1,0);
printf("%d\n",tot);
for (int i = 1;i<=tot;i++)
{
printf("%d ",ans[i].size());
for (int j = 0;j<ans[i].size();j++)
printf("%d ",ans[i][j]);
printf("\n");
}
}
Description
In Berland there are n cities and n - 1
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to work simultaneously for one day. Both brigades repair one road for the whole day and cannot take part in repairing other roads on that day. But the repair brigade can do nothing on that day.
Determine the minimum number of days needed to repair all the roads. The brigades cannot change the cities where they initially are.
Input
The first line of the input contains a positive integer n (2 ≤ n ≤ 200 000) — the number of cities in Berland.
Each of the next n - 1 lines contains two numbers ui, vi, meaning that the i-th road connects city ui and city vi (1 ≤ ui, vi ≤ n, ui ≠ vi).
Output
First print number k
In next k lines print the description of the roads that should be repaired on each of the k days. On the i-th line print first number di — the number of roads that should be repaired on the i-th day, and then di space-separated integers — the numbers of the roads that should be repaired on the i-th day. The roads are numbered according to the order in the input, starting from one.
If there are multiple variants, you can print any of them.
Sample Input
Input
4 1 2 3 4 3 2
Output
2 2 2 1 1 3
Input
6 3 4 5 4 3 2 1 3 4 6
Output
3 1 1 2 2 3 2 4 5
