E - Dominant Indices

最新推荐文章于 2022-04-01 10:30:34 发布

原创最新推荐文章于 2022-04-01 10:30:34 发布 · 229 阅读

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本文详细解析了一种基于树形结构的算法问题，旨在计算每个节点的深度数组及其主导索引。通过递归深度优先搜索（DFS）策略，结合map数据结构来记录每个节点子树中各个深度的节点数，最终确定每个节点的主导深度。代码实现采用C++，展示了如何高效处理大规模树形数据。

You are given a rooted undirected tree consisting of n vertices. Vertex 1 is the root.

Let’s denote a depth array of vertex x as an infinite sequence [dx,0,dx,1,dx,2,…], where dx,i is the number of vertices y such that both conditions hold:

x is an ancestor of y;
the simple path from x to y traverses exactly i edges.
The dominant index of a depth array of vertex x (or, shortly, the dominant index of vertex x) is an index j such that:

for every k<j, dx,k<dx,j;
for every k>j, dx,k≤dx,j.
For every vertex in the tree calculate its dominant index.

Input
The first line contains one integer n (1≤n≤106) — the number of vertices in a tree.

Then n−1 lines follow, each containing two integers x and y (1≤x,y≤n, x≠y). This line denotes an edge of the tree.

It is guaranteed that these edges form a tree.

Output
Output n numbers. i-th number should be equal to the dominant index of vertex i.

Examples
Input
4
1 2
2 3
3 4
Output
0
0
0
0
Input
4
1 2
1 3
1 4
Output
1
0
0
0
Input
4
1 2
2 3
2 4
Output
2
1
0
0
题意变化一下就是找到每个结点子树中同一深度最多的那个深度，相同时取最近的，用map记录每个结点子树结点在每个深度下的数量，用两个数组分别记录深度最多的结点数量和最近深度用来更新。

#include<bits/stdc++.h>
using namespace std;
const int N=1e6+5;
int rt[N],dep[N];
int ma[N],num[N];
int ans[N];
vector<int> edge[N];
map<int,int> m[N];
map<int,int>::iterator it;

int merge(int x,int y)
{
	if(m[x].size()<m[y].size())
	swap(x,y);
	
	for(it=m[y].begin();it!=m[y].end();it++)
	{
		m[x][it->first]+=it->second;
		if(m[x][it->first]>num[x]||(m[x][it->first]==num[x]&&(it->first)<=ma[x]))
		{
			num[x]=m[x][it->first];
			ma[x]=it->first;
		}
	}
	return x;
}

void dfs(int x,int fa)
{
	dep[x]=dep[fa]+1;
	ma[rt[x]]=dep[x];
	num[rt[x]]=1;
	m[rt[x]][dep[x]]++;
	for(int i=0;i<edge[x].size();i++)
	if(edge[x][i]!=fa)
	dfs(edge[x][i],x);
	
	for(int i=0;i<edge[x].size();i++)
	{
		if(edge[x][i]!=fa)
		rt[x]=merge(rt[x],rt[edge[x][i]]);
	}
	ans[x]=ma[rt[x]];
}

int n,a,b;
int main()
{
	scanf("%d",&n);
	dep[0]=0;
	for(int i=1;i<n;i++)
	{
		scanf("%d%d",&a,&b);
		edge[a].push_back(b);
		edge[b].push_back(a);
		rt[i]=i;
	}
	rt[n]=n;
	dfs(1,0);
	for(int i=1;i<=n;i++)
	printf("%d\n",ans[i]-dep[i]);
}