Codeforces734E【縮點+樹的直徑】

思路：
想不到樹的直徑就GG了啊…
先縮點，然後縮成一棵相鄰兩層節點就是不同的個數，然後如果考慮一棵樹是

1-0-1-0-1-0-1-0-1-0

這樣的話其實就是中間開始搞，然後最少變node_num/2次變成同一種顏色。

然後考慮一棵複雜的樹，當樹上最長距離像這樣在變的時候其餘枝幹肯定是會被連帶到的！

//#pragma comment(linker, "/STACK:102400000,102400000")
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std;
typedef long long LL;
#define mem(a, b) memset(a, b, sizeof(a))

const int Maxn = 2e5 + 10;
struct Edge{
    int v, nex;
}edge[Maxn<<1], e[Maxn<<1];
int head[Maxn], tol, op[Maxn], tal;
int val[Maxn], tot, col[Maxn], be[Maxn];
int n;

void init(){
    tol = tot = tal = 0;
    memset(op, -1, sizeof(op));
    memset(head, -1, sizeof(head));
}

void add(int u, int v){
    edge[tol] = (Edge){v, head[u]}, head[u] = tol++;
    edge[tol] = (Edge){u, head[v]}, head[v] = tol++;
}
void ADD(int u, int v){
    e[tal] = (Edge){v, op[u]}, op[u] = tal++;
    e[tal] = (Edge){u, op[v]}, op[v] = tal++;
}
void DFS(int u, int fa){
    int v;
    if(fa != -1){
        if(col[be[fa]] != val[u]){
            be[u] = ++tot;
            col[be[u]] = val[u];
            ADD(be[fa], tot);
        }
        else be[u] = be[fa];
    }
    for(int i=head[u]; ~i; i = edge[i].nex){
        v = edge[i].v;
        if(v == fa) continue;
        DFS(v, u);
    }
}

int Mmmax(int x, int y, int z){
    x = x > y ? x : y;
    return x > z ? x : z;
}
int dis[Maxn];
int ans;
void DFS_LEN(int u, int fa, int pre_num){
    int v;
    int temp, Max1, Max2;
    Max1 = Max2 = 0;
    dis[u] = 1;
    for(int i=op[u]; ~i; i=e[i].nex){
        v = e[i].v;
        if(v == fa) continue;
        temp = Mmmax(Max1, Max2, pre_num);
        DFS_LEN(v, u, temp);
        if(dis[v] > Max1){
            Max2 = Max1;
            Max1 = dis[v];
        }
        else if(dis[v] > Max2) Max2 = dis[v];
    }
    dis[u] = dis[u] + Max1;
    ans = Mmmax(ans, Max1+Max2, Max1+pre_num);
}

int main(){
    int u, v;
    scanf("%d", &n);
    for(int i=1;i<=n;i++) scanf("%d", &val[i]);
    init();
    for(int i=1;i<n;i++){
        scanf("%d%d", &u, &v);
        add(u, v);
    }
    col[++tot] = val[1];
    be[1] = tot;
    DFS(1, -1);
    ans = 0;
    DFS_LEN(1, -1, 0);
    printf("%d\n", (ans+1)/2);
    return 0;
}