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

题意：给你一棵树， 每个点要么是黑色要么是白色， 有一种操作是将同一个颜色的连通块变成相反的颜色，问你最少变换几次， 整颗树变成一种颜色。

 

思路： 缩点， 加求树的直径， 答案为树的直径除二向上取整。

 

#include<bits/stdc++.h>
#define LL long long
#define mk make_pair
using namespace std;

const int N = 5e5 + 7;
const int inf = 0x3f3f3f3f;
const int INF = 0x3f3f3f3f3f3f3f3f;

int n, a[N], cnt[N], tot, vis[N], depth[N], id, ans;
vector<int> edge1[N], edge2[N];

void dfs(int u, int pre, int op, int f) {
    vis[u] = f;
    for(int v : edge1[u]) {
        if(v == pre || a[v] != op)
            continue;
        dfs(v, u, op, f);
    }
}

void dfs2(int u, int pre, int deep) {
    depth[u] = deep;
    if(depth[u] > depth[id])
        id = u;
    for(int v : edge2[u]) {
        if(v == pre) continue;
        dfs2(v, u, deep + 1);
    }
}

void dfs3(int u, int pre, int deep) {
    ans = max(ans, deep);
    for(int v : edge2[u]) {
        if(v == pre) continue;
        dfs3(v, u, deep + 1);
    }
}
int main() {
    scanf("%d", &n);
    for(int i = 1; i <= n; i++) {
        scanf("%d", &a[i]);
    }

    for(int i = 1; i < n; i++) {
        int u, v; scanf("%d%d", &u, &v);
        edge1[u].push_back(v);
        edge1[v].push_back(u);
    }

    for(int i = 1; i <= n; i++) {
        if(vis[i]) continue;    
        dfs(i, 0, a[i], ++tot);
    }

    for(int u = 1; u <= n; u++) {
        for(int v : edge1[u]) {
            if(u > v || vis[u] == vis[v])
                continue;
            edge2[vis[u]].push_back(vis[v]);
            edge2[vis[v]].push_back(vis[u]);
        }
    }

    dfs2(1, 0, 0);
    dfs3(id, 0, 0);
    if(ans & 1) ans++;
    printf("%d\n", ans / 2);
    return 0;
}
/*
*/