CF 219 D Choosing Capital for Treeland（树形dp）

D. Choosing Capital for Treeland

time limit per test
3 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

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.

Sample test(s)

input

3
2 1
2 3

output

0
2

input

4
1 4
2 4
3 4

output

2
1 2 3

/*
这题就是给你一棵有向树，需要选定一个点为capital，满足翻转边数最小。
思路:先求出1为capital 的答案，然后向下更新孩子节点

*/

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<set>
#include<map>

#define L(x) (x<<1)
#define R(x) (x<<1|1)
#define MID(x,y) ((x+y)>>1)

#define bug printf("hihi\n")

#define eps 1e-8
typedef __int64 ll;

using namespace std;
#define INF 0x3f3f3f3f
#define N 200005

int dp
;
int n;

struct stud{
  int to,ne,len;
}e[N*2];
int head
;
int num;

inline void add(int u,int v,int len)
{
    e[num].to=v;
    e[num].len=len;
    e[num].ne=head[u];
    head[u]=num++;
}

void dfs(int u,int pre)
{
    dp[u]=0;
    for(int i=head[u];i!=-1;i=e[i].ne)
    {
        int to=e[i].to;
        if(to==pre) continue;
        dfs(to,u);
        dp[u]+=dp[to]+e[i].len;
    }
}

void dfsup(int u,int pre)
{
    for(int i=head[u];i!=-1;i=e[i].ne)
    {
        int to=e[i].to;
        if(to==pre) continue;
        dp[to]=dp[u];
        if(e[i].len) dp[to]--;
        else dp[to]++;
        dfsup(to,u);
    }
}

int main()
{
    int i,j;
    while(~scanf("%d",&n))
    {
        memset(head,-1,sizeof(head));
        num=0;
        int u,v;
        i=n-1;
        while(i--)
        {
            scanf("%d%d",&u,&v);
            add(u,v,0);
            add(v,u,1);
        }
        dfs(1,-1);
        dfsup(1,-1);
        int ans=INF;
        for(i=1;i<=n;i++)
            ans=min(ans,dp[i]);
        printf("%d\n",ans);
        vector<int>temp;
        temp.clear();
        for(i=1;i<=n;i++)
            if(dp[i]==ans) temp.push_back(i);
        for(i=0;i<temp.size();i++)
        {
            if(i) printf(" ");
            printf("%d",temp[i]);
        }
        printf("\n");
    }
    return 0;
}