Codeforces Round #379 (Div. 2) E Anton and Tree（缩点）

思路：一个相同颜色的联通块可以把颜色反转，那么容易想到把这一个联通块缩点成一块，那么最后得出来的就是一颗黑白相间的树，那么最后的结果就是直径/2向上取整啦

#include<bits/stdc++.h>
using namespace std;
#define LL long long
const int maxn = 2*1e5+7;
int c[maxn],d[maxn];
vector<int>e[maxn];
void dfs(int u,int fa)
{
for(int i =0;i<e[u].size();i++)
{
int v= e[u][i];
if(v==fa)continue;
d[v]=d[u]+(c[u]^c[v]);
dfs(v,u);
}
}
int main()
{
int n;
scanf("%d",&n);
for(int i = 1;i<=n;i++)scanf("%d",&c[i]);
for(int i = 1;i<=n-1;i++)
{
int u,v;
scanf("%d%d",&u,&v);
e[u].push_back(v);
e[v].push_back(u);
}
int u = 1;
dfs(u,-1);
int maxlen = d[u];
for(int i = 1;i<=n;i++)
{
if(d[i]>maxlen)
{
maxlen = d[i];
u = i;
}
}
memset(d,0,sizeof(d));
dfs(u,-1);
maxlen = d[u];
for(int i=1;i<=n;i++)
{
if(d[i]>maxlen)
{
maxlen = d[i];
u=i;
}
}
printf("%d\n",(d[u]+1)/2);
}

E. Anton and Tree

time limit per test
3 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

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



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



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 01 22 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.