VK Cup 2016 - Qualification Round 2 C. Road Improvement dfs

C. Road Improvement

题目连接：

http://www.codeforces.com/contest/638/problem/C

Description

In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.

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 — the minimum number of days needed to repair all the roads in Berland.

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

4
1 2
3 4
3 2

Sample Output

2
2 2 1
1 3

Hint

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。

题解：

直接dfs就好了，我们记录一个上一个边传过来的时间和这条边修理的时间就好了

保证不重复就行了。

代码

#include<bits/stdc++.h>
using namespace std;
const int maxn = 2e5+7;
vector<pair<int,int> >E[maxn];
vector<int>ans[maxn];
int tot = 0;
void dfs(int x,int fa,int te)
{
    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==te)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 x,y;scanf("%d%d",&x,&y);
        E[x].push_back(make_pair(y,i));
        E[y].push_back(make_pair(x,i));
    }
    dfs(1,-1,0);
    cout<<tot<<endl;
    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");
    }
}