Codeforces Round #135 (Div. 2)VD. 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 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 

題意：給你一那個點，n-1條變的單項路，讓你找出一個點能到達其他所有點 所需要反向道路的最小條數並輸出起點。

解題思路： 首先找一個點最爲起點，算出以他爲起點所需要反向道路的條數，再用此起點更新其他點（i能連到j,dp[j]=dp[i]+1,否則dp[j]=dp[i]-1）

代碼：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define N 200005
int ans;
int dp[N];
struct node
{
    int st,en,c,next;
}e[2*N];
int p[N],num;
void init()
{
    memset(p,-1,sizeof(p));
    num=0;
}
void add(int st,int en,int c)
{
    e[num].st=st;
    e[num].en=en;
    e[num].c=c;
    e[num].next=p[st];
    p[st]=num++;
}
void dfs1(int root,int fa)
{
    for(int i=p[root];i+1;i=e[i].next)
    {
        int son=e[i].en;
        if(fa==son)continue;
        dfs1(son,root);
        if(e[i].c==0)
        {
            ans+=1;
        }
        else
            ans=ans+0;
    }
}
void dfs2(int root,int fa)
{
    for(int i=p[root];i+1;i=e[i].next)
    {
        int son=e[i].en;
        if(son==fa)continue;
        if(e[i].c==0)
            dp[son]=dp[root]-1;
        else
            dp[son]=dp[root]+1;
        dfs2(son,root);
    }
}
int main()
{
    int n,a,b;
    while(scanf("%d",&n)!=EOF)
    {
        memset(dp,0,sizeof(dp));
        init();
        for(int i=1;i<n;i++)
        {
            scanf("%d%d",&a,&b);
            add(a,b,1);
            add(b,a,0);
        }

        ans=0;
        dfs1(1,-1);
        dp[1]=ans;
        dfs2(1,-1);
       int ans1=99999999;
       for(int i=1;i<=n;i++)
       {
           if(ans1>dp[i])
            ans1=dp[i];
       }
       printf("%d\n",ans1);
       for(int i=1;i<=n;i++)
       {
           if(dp[i]==ans1)
           printf("%d ",i);
       }
       printf("\n");
    }
    return 0;
}