Codeforces-734E Anton and Tree（树的直径）

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 andu). 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

110 0 0 1 1 0 1 0 0 1 11 21 32 42 55 65 73 83 93 109 11

output

2

input

40 0 0 01 22 33 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.

题解：把所有颜色相同且相连的点缩成一个点，然后重新构建树，可以发现黑白是相隔出现的，只要找出树的直径，把直径中心的点连续变换len/2次即可让所有的节点颜色相同，len为直径长度

#include<bits/stdc++.h>#define x first#define y secondusing namespace std;typedef pair<int, int> PII;typedef long long LL;const int MX = 2e5 + 5;const int inf = 0x3f3f3f3f;struct Edge {    int v, nxt;} E[MX * 2];int head[MX], tot;void init() {    memset(head, -1, sizeof(head));    tot = 0;}void add(int u, int v) {    E[tot].v = v;    E[tot].nxt = head[u];    head[u] = tot++;}int n, vis[MX], c[MX], p[MX];void dfs(int u, int rt) {    vis[u] = 1;    p[u] = rt;    for (int i = head[u]; ~i; i = E[i].nxt) {        int v = E[i].v;        if (c[v] != c[u] || vis[v]) continue;        dfs(v, rt);    }}int dep[MX];void DFS(int u, int fa) {    for (int i = head[u]; ~i; i = E[i].nxt) {        int v = E[i] .v;        if (v == fa) continue;        dep[v] = dep[u] + 1;        DFS(v, u);    }}PII e[MX];int main() {    //freopen("in.txt", "r", stdin);    init();    scanf("%d", &n);    for (int i = 1; i <= n; i++) scanf("%d", &c[i]);    for (int i = 1; i < n; i++) {        scanf("%d%d", &e[i].x, &e[i].y);        add(e[i].x, e[i].y); add(e[i].y, e[i].x);    }    memset(vis,0,sizeof(vis));    for (int i = 1; i <= n; i++) if (!vis[i]) dfs(i, i);    init();    for (int i = 1; i < n; i++) {        if (p[e[i].x] != p[e[i].y]) {            add(p[e[i].x], p[e[i].y]);            add(p[e[i].y], p[e[i].x]);        }    }    int rt = p[1];    dep[rt] = 1;    DFS(rt, -1);    for (int i = 1; i <= n; i++) if (p[i] == i && dep[i] > dep[rt]) rt = i;    int mx = 0;    dep[rt] = 1;    DFS(rt, -1);    for (int i = 1; i <= n; i++) if (p[i] == i && dep[i] > mx) mx = dep[i];    //printf("%d\n",mx);    printf("%d\n", mx / 2);    return 0;}