CF734E：Anton and Tree（缩点 & 直径）

最新推荐文章于 2019-05-24 08:38:23 发布

原创最新推荐文章于 2019-05-24 08:38:23 发布 · 532 阅读

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探讨在一棵带有两种颜色的树形图中，通过特定操作使得所有节点颜色统一所需的最少步骤。文章介绍了一种有效的算法，通过节点压缩并求解树的直径来确定最优解。

Anton is growing a tree in his garden. In case you forgot, the tree is a connected acyclic undirected graph.

There are n vertices in the tree, each of them is painted black or white. Anton doesn't like multicolored trees, so he wants to change the tree such that all vertices have the same color (black or white).

To change the colors Anton can use only operations of one type. We denote it as paint(v), where v is some vertex of the tree. This operation changes the color of all vertices u such that all vertices on the shortest path from v to u have the same color (including v and u). For example, consider the tree

$\text{[math]}$

and apply operation paint(3) to get the following:

$\text{[math]}$

Anton is interested in the minimum number of operation he needs to perform in order to make the colors of all vertices equal.

Input

The first line of the input contains a single integer n (1 ≤ n ≤ 200 000) — the number of vertices in the tree.

The second line contains n integers colori (0 ≤ colori ≤ 1) — colors of the vertices. colori = 0 means that the i-th vertex is initially painted white, while colori = 1 means it's initially painted black.

Then follow n - 1 line, each of them contains a pair of integers ui and vi (1 ≤ ui, vi ≤ n, ui ≠ vi) — indices of vertices connected by the corresponding edge. It's guaranteed that all pairs (ui, vi) are distinct, i.e. there are no multiple edges.

Output

Print one integer — the minimum number of operations Anton has to apply in order to make all vertices of the tree black or all vertices of the tree white.

Examples

input
11
0 0 0 1 1 0 1 0 0 1 1
1 2
1 3
2 4
2 5
5 6
5 7
3 8
3 9
3 10
9 11

output
2

input
4
0 0 0 0
1 2
2 3
3 4

output
0

Note

In the first sample, the tree is the same as on the picture. If we first apply operation paint(3) and then apply paint(6), the tree will become completely black, so the answer is 2.

In the second sample, the tree is already white, so there is no need to apply any operations and the answer is 0.

题意：一棵树有两种颜色，每次操作可以将一个节点及其所在的联通块变色，问使整棵树变成同一种色要几次操作。

思路：缩点，求树的直径d，(d+1)/2就是答案，即从最长链的中间开始变色。

# include <bits/stdc++.h>
# define pb push_back
using namespace std;
const int maxn = 2e5+30;
vector<int>v[maxn], s[maxn];
int ans=0, a[maxn], cnt=0, vis[maxn], h[maxn], fa[maxn];
void dfs(int cur, int tag)
{
    vis[cur] = 1;
    fa[cur] = tag;
    for(auto to : v[cur])
    {
        if(vis[to] || a[to] != a[cur] ) continue;
        vis[to] = 1;
        dfs(to, tag);
    }
}

void dfs2(int cur, int pre)
{
    int mx=0, mmx=0;
    for(auto to : s[cur])
    {
        if(to == pre) continue;
        dfs2(to, cur);
        mx = max(mx, h[to]+1);
        if(mx > mmx) swap(mx, mmx);//维护两个最深的儿子。
    }
    h[cur] = mmx;
    ans = max(ans, mx+mmx);
}
int main()
{
    int n;
    scanf("%d",&n);
    for(int i=1; i<=n; ++i) scanf("%d",&a[i]);
    for(int i=1; i<n; ++i)
    {
        int a, b;
        scanf("%d%d",&a,&b);
        v[a].pb(b), v[b].pb(a);
    }
    for(int i=1; i<=n; ++i)
        if(!vis[i])
            dfs(i, ++cnt);
    for(int i=1; i<=n; ++i)
        for(auto j : v[i])
            if(a[j] != a[i]) s[fa[i]].pb(fa[j]);
    dfs2(1, 0);
    printf("%d\n",(ans+1)/2);
    return 0;
}