We have a tree $T$ with vertices numbered $1$ to $N$. The $i$-th edge of $T$ connects vertex $u_i$ and vertex $v_i$. Let us use $T$ to define the similarity of a permutation $P = (P_1,P_2,\ldots,P_N)$ of $(1,2,\ldots,N)$ as follows. Construct a permutation $P$ with the minimum similarity. A subsequence of a sequence is a sequence obtained by removing zero or more elements from that sequence and concatenating the remaining elements without changing the relative order.Problem Statement
What is a subsequence?
For instance, \((10,30)\) is a subsequence of \((10,20,30)\), but \((20,10)\) is not.What is a simple path?
For vertices $X$ and $Y$ in a graph $G$, a walk from $X$ to $Y$ is a sequence of vertices $v_1,v_2, \ldots, v_k$ such that $v_1=X$, $v_k=Y$, and there is an edge connecting $v_i$ and $v_{i+1}$. A simple path (or simply a path) is a walk such that $v_1,v_2, \ldots, v_k$ are all different.
Constraints
Input
The input is given from Standard Input in the following format:
$N$
$u_1$ $v_1$
$u_2$ $v_2$
$\vdots$
$u_{N-1}$ $v_{N-1}$
Output
Print a permutation $P$ with the minimum similarity, separated by spaces. If multiple solutions exist, you may print any of them.
Sample Input 1
3 1 2 2 3
Sample Output 1
3 2 1
This permutation has a similarity of $1$, which can be computed as follows.
-
For $x=(1)$, we have $y=(P_1)=(3)$. The length of a longest common subsequence of $x$ and $y$ is $0$.
-
For $x=(2)$, we have $y=(P_2)=(2)$. The length of a longest common subsequence of $x$ and $y$ is $1$.
-
For $x=(3)$, we have $y=(P_2)=(1)$. The length of a longest common subsequence of $x$ and $y$ is $0$.
-
For $x=(1,2)$, we have $y=(P_1,P_2)=(3,2)$. The length of a longest common subsequence of $x$ and $y$ is $1$. The same goes for $x=(2,1)$, the reversal of $(1,2)$.
-
For $x=(2,3)$, we have $y=(P_2,P_3)=(2,1)$. The length of a longest common subsequence of $x$ and $y$ is $1$. The same goes for $x=(3,2)$, the reversal of $(2,3)$.
-
For $x=(1,2,3)$, we have $y=(P_1,P_2,P_3)=(3,2,1)$. The length of a longest common subsequence of $x$ and $y$ is $1$. The same goes for $x=(3,2,1)$, the reversal of $(1,2,3)$.
We can prove that no permutation has a similarity of $0$ or less, so this permutation is a valid answer.
首先看样例很容易知道，答案至多为1.
那么想要使两个串的 LCS 为1，可以尝试不断剖叶子构造。
维护一个叶子队列，每次取出两个叶子 \(a,b\)，令 \(P_a=b,P_b=a\)。这样构造是合法的。
首先假设证明了之前的链的的 LCS 都是 1，此时加入这样两个叶子，那么如果想要让 LCS 更大，需要使选的序列在 \(a,b\) 的后面，那么此时 LCS 仍是1.
#include<bits/stdc++.h>
using namespace std;
const int N=5005;
int fa[N],in[N],l=1,r=0,q[N],n,rt,hd[N],e_num,p[N];
struct edge{
int v,nxt;
}e[N<<1];
void add_edge(int u,int v)
{
e[++e_num]=(edge){v,hd[u]};
hd[u]=e_num;
}
int main()
{
scanf("%d",&n);
if(n==2)
{
puts("2 1");
return 0;
}
for(int i=1,u,v;i<n;i++)
scanf("%d%d",&u,&v),in[u]++,in[v]++,add_edge(u,v),add_edge(v,u);
for(int i=1;i<=n;i++)
if(in[i]==1)
rt=i,q[++r]=i;;
while(l<r)
{
p[q[l]]=q[l+1];
p[q[l+1]]=q[l];
for(int i=hd[q[l]];i;i=e[i].nxt)
{
in[e[i].v]--;
if(in[e[i].v]==1)
q[++r]=e[i].v;
}
for(int i=hd[q[l+1]];i;i=e[i].nxt)
{
in[e[i].v]--;
if(in[e[i].v]==1)
q[++r]=e[i].v;
}
l+=2;
}
// printf("%d %d\n",l,r);
if(l==r)
p[q[l]]=q[l];
for(int i=1;i<=n;i++)
printf("%d ",p[i]);
return 0;
}