CodeForces 219D Choosing Capital for Treeland

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 s_i, t_i (1 ≤ s_i, t_i ≤ n; s_i ≠ t_i) — the numbers of cities, connected by that road. The i-th road is oriented from city s_i to city t_i. 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

32 12 3

Output

02 

Input

41 42 43 4

Output

21 2 3 

#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>#include <vector>#include <string>#include <queue>#include <ctime>#include <cstdlib>#include <stack>#include <map>#include <set>#include <list>#define MP make_pair#define PB push_back#define INT_INF 0x3fffffff#define LL_INF 0x3fffffffffffffff#define EPS 1e-12#define MOD 1000000007#define PI 3.14159265358979323846#define N 200010#define E 400010using namespace std;typedef long long LL;typedef unsigned long long ULL;typedef unsigned int Uint;typedef double DB;struct Edge{    int st,en,val,next;} edge[E];int head[N] , tot;int f[N];void add_edge(int st,int en,int val){    edge[tot].st=st;    edge[tot].en=en;    edge[tot].val=val;    edge[tot].next=head[st];    head[st]=tot++;}void dfs1(int u,int pre){    f[u]=0;    for(int e=head[u]; e!=-1; e=edge[e].next)    {        int v=edge[e].en;        if(v==pre) continue;        f[u]+=edge[e].val;        dfs1(v,u);        f[u]+=f[v];    }}vector<int> res;int ans;void dfs2(int u,int pre,int now){    if(now+f[u]<ans)    {        res.clear();        ans=now+f[u];        res.PB(u);    }    else if(now+f[u]==ans) res.PB(u);    for(int e=head[u]; e!=-1; e=edge[e].next)    {        int v=edge[e].en;        if(v==pre) continue;        int cnt=now+f[u]-f[v];        if(edge[e].val==0) cnt++;        if(edge[e].val==1) cnt--;        dfs2(v,u,cnt);    }}int main(){    int n;    while(~scanf("%d",&n))    {        memset(head,-1,sizeof(head));        tot=0;        for(int i=1,a,b; i<n; i++)        {            scanf("%d%d",&a,&b);            add_edge(a,b,0);            add_edge(b,a,1);        }        dfs1(1,-1);        ans=INT_INF;        res.clear();        dfs2(1,-1,0);        sort(res.begin(),res.end());        printf("%d\n",ans);        for(int i=0; i<(int)res.size(); i++)            printf("%d ",res[i]);        printf("\n");    }    return 0;}