Choosing Capital for Treeland （树形dp+双向搜索）

题目：Choosing Capital for Treeland 

The country Treeland consists of n cities, some pairs of them are connected with unidirectional roads. Overall there are n - 1 roads in the country. We know that if we don't take the direction of the roads into consideration, we can get from any city to any other one.

The council of the elders has recently decided to choose the capital of Treeland. Of course it should be a city of this country. The council is supposed to meet in the capital and regularly move from the capital to other cities (at this stage nobody is thinking about getting back to the capital from these cities). For that reason if city a is chosen a capital, then all roads must be oriented so that if we move along them, we can get from city a to any other city. For that some roads may have to be inversed.

Help the elders to choose the capital so that they have to inverse the minimum number of roads in the country.

Input
The first input line contains integer n (2 ≤ n ≤ 2·105) — the number of cities in Treeland. Next n - 1 lines contain the descriptions of the roads, one road per line. A road is described by a pair of integers si, ti (1 ≤ si, ti ≤ n; si ≠ ti) — the numbers of cities, connected by that road. The i-th road is oriented from city si to city ti. You can consider cities in Treeland indexed from 1 to n.

Output
In the first line print the minimum number of roads to be inversed if the capital is chosen optimally. In the second line print all possible ways to choose the capital — a sequence of indexes of cities in the increasing order.

Example
Input
3
2 1
2 3
Output
0
2 
Input
4
1 4
2 4
3 4
Output
2
1 2 3 

题意：

有一棵n个结点的树，有n-1条有向边，每条边的方向都可以翻转，翻转的代价是1，问从任意一个节点出发，到其他所有节点要逆向走多少条路的最小值。

思路：

题目的树是一棵有方向的树，所以对于任意一个节点，搜索向上和向下两个方向。

用0和1表示边的方向，题目就转换为了求每个点到其他点的权值问题。

设置一个dp数组。

dp[i][0]表示以i为根的子树中，从i出发，要反转多少条边

dp[i][1]表示从i往父亲方向走，要反转多少条边

这一类的题目，构造好树形结构，找父亲和孩子的关系，

向下遍历，向上更新，先dfs后dp

向下遍历，向下更新，先dp后dfs

源代码：