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.
Examples
Input
32 12 3
Output
02
Input
41 42 43 4
Output
21 2 3
/*
纪念自己第一道完全自己想出来的树DP！！！！
给n个城市，n-1单向道路，要从中选择一个首都，首都的要求是从首都能到达所有其他城市，如果这些单向路中的方向不合适，可以重修这条路；
问选择一个最合适的首都，是的修路的次数最少；
思路：建图时，将路标记正向为0，逆向为1，dp[u]表示以u为首都最少修的路，dfs出dp[1]然后就能得出一个状态转移方程dp[v]=dp[u]+(w?-1:1);
再用一次dfs就可以了
*/
#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<string.h>
#include<vector>
#define N 200010
#define INF 0x3f3f3f3f
using namespace std;
struct node
{
int to,len;//下一个节点，长度
node (int x,int y)
{
to=x;
len=y;
}
};
int n;
vector<node> edge[N*2];
int dp[N];
int ans=0;
int dfs1(int u,int p)
{
for(int i=0;i<edge[u].size();i++)
{
int v=edge[u][i].to;
int w=edge[u][i].len;
if(v==p) continue;
//cout<<"u="<<u<<" "<<"v="<<v<<" "<<"w="<<w<<endl;
ans+=w;
dfs1(v,u);
}
return ans;
}
void dfs2(int u,int p)
{
for(int i=0;i<edge[u].size();i++)
{
int v=edge[u][i].to;
int w=edge[u][i].len;
if(v==p) continue;
//cout<<"u="<<u<<" "<<"v="<<v<<" "<<"w="<<w<<endl;
dp[v]=dp[u]+(w?-1:1);
dfs2(v,u);
}
}
int main()
{
//freopen("in.txt","r",stdin);
while(scanf("%d",&n)!=EOF)
{
memset(dp,0,sizeof dp);
for(int i=0;i<=n;i++)
edge[i].clear();
int a,b;
for(int i=1;i<=n-1;i++)
{
scanf("%d%d",&a,&b);
edge[a].push_back(node(b,0));
edge[b].push_back(node(a,1));//逆向走的话就得花费一次
}
int min=INF;
ans=0;
dp[1]=dfs1(1,-1);
//cout<<"dp[1]="<<dp[1]<<endl;
dfs2(1,-1);
for(int i=1;i<=n;i++)
{
if(dp[i]<min)
min=dp[i];
}
printf("%d\n",min);
int flag=0;
for(int i=1;i<=n;i++)
{
if(dp[i]==min)
{
printf(flag?" %d":"%d",i);
flag=1;
}
}
printf("\n");
}
return 0;
}