AIM Tech Round 3 (Div. 1) C. Centroids(树形DP)

C. Centroids

time limit per test
4 seconds

memory limit per test
512 megabytes

input
standard input

output
standard output

Tree is a connected acyclic graph. Suppose you are given a tree consisting ofn vertices. The vertex of this tree is calledcentroid if the
size of each connected component that appears if this vertex is removed from the tree doesn't exceed

.

You are given a tree of size n and can perform no more than one edge replacement.Edge replacement is the operation of removing one edge from the tree (without deleting incident
vertices) and inserting one new edge (without adding new vertices) in such a way that the graph remains a tree. For each vertex you have to determine if it's possible to make it centroid by performing no more than one edge replacement.

Input
The first line of the input contains an integer n (2 ≤ n ≤ 400 000) — the number of vertices in the tree. Each of the nextn - 1 lines contains
a pair of vertex indicesui andvi (1 ≤ ui, vi ≤ n) —
endpoints of the corresponding edge.

Output
Print n integers. The
i-th of them should be equal to 1 if thei-th vertex can be made centroid by replacing no more than one edge, and should be equal to0 otherwise.

Examples

Input

3
1 2
2 3

Output

1 1 1

Input

5
1 2
1 3
1 4
1 5

Output

1 0 0 0 0

Note
In the first sample each vertex can be made a centroid. For example, in order to turn vertex1 to centroid one have to replace the edge
(2, 3) with the edge (1, 3).

题意：给你一棵树，对于每个顶点，问最多更改一条边后能否把它变为树的重心。

分析：树形DP，计算向下和向上最多能找到的点数不大于n/2的子树，注意如果一个点不是重心，那么它最多只有一个儿子的节点数超过n/2,我们计算完所有辅助值后用改超标儿子的度数减去它的最大不超标子树，判断此时的值是否还大于n/2即可。对于这种树形DP记录最优解和次优解辅助转移是一种常见技巧。

#include<iostream>
#include<string>
#include<algorithm>
#include<cstdlib>
#include<cstdio>
#include<set>
#include<map>
#include<vector>
#include<cstring>
#include<stack>
#include<cmath>
#include<queue>
#include <unordered_map>
#define INF 0x3f3f3f3f
#define eps 1e-9
#define MOD 1000000007
#define MAXN 400005
using namespace std;
typedef long long ll;
vector<int> G[MAXN];
int n,sz[MAXN],up[MAXN],Down[MAXN],max1[MAXN],max2[MAXN],ans[MAXN];
void update(int val,int &x,int &y)
{
if(y < val) y = val;
if(x < y) swap(x,y);
}
void dfs1(int u,int fa)
{
sz[u] = 1;
for(int v : G[u])
if(v != fa)
{
dfs1(v,u);
sz[u] += sz[v];
update(Down[v],max1[u],max2[u]);
}
Down[u] = max1[u];
if(sz[u] <= n/2) Down[u] = sz[u];
}
void dfs2(int u,int fa)
{
if(fa >= 1)
{
up[u] = max(up[u],up[fa]);
if(Down[u] != max1[fa]) up[u] = max(up[u],max1[fa]);
else up[u] = max(up[u],max2[fa]);
if(n - sz[u] <= n/2) up[u] = n - sz[u];
}
int tmp = 0;
for(int v : G[u])
if(v != fa)
{
dfs2(v,u);
if(sz[v] > n/2) tmp = v;
}
if(n - sz[u] > n/2) tmp = -1;
if(!tmp) return;
if(tmp > 0)
{
if(sz[tmp] - Down[tmp] > n/2) ans[u] = 1;
return;
}
if(n - sz[u] - up[u] > n/2) ans[u] = 1;
}
int main()
{
scanf("%d",&n);
for(int i = 1;i < n;i++)
{
int x,y;
scanf("%d%d",&x,&y);
G[x].push_back(y);
G[y].push_back(x);
}
dfs1(1,-1);
dfs2(1,-1);
for(int i = 1;i <= n;i++)
printf("%d ",ans[i]^1);
}