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题目翻译

给一棵n个节点的树, 节点编号为1~n。

树上的结点要么是黑色，要么是白色，每次可以把连通的一种颜色变成另一种颜色。求至少要多少次，才能是整棵树变为一种颜色。

思路

不难发现，如果我们把两个相邻的点是否相同作为边权，这道题目其实就是求出这棵树的直径上有多少种成对的不同的颜色。

代码

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/**
 * @file CF734E.cpp
 * @author Zhoumouren
 * @brief 
 * @version 0.1
 * @date 2024-08-22
 * 
 * @copyright Copyright (c) 2024

easy , just count the different color's pair on the diameter of tree.

*/

#pragma GCC optimize(2)
#pragma GCC optimize(3, "Ofast", "inline")
#include<bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef __int128 int128;

namespace FastIO 
{
    char buf[1 << 20], *p1, *p2;
    #define getchar() (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 20, stdin), p1 == p2) ? EOF : *p1++)
    template<typename T> inline T read(T& x) {
        x = 0;
        int f = 1;
        char ch;
        while (!isdigit(ch = getchar())) if (ch == '-') f = -1;
        while (isdigit(ch)) x = (x << 1) + (x << 3) + (ch ^ 48), ch = getchar();
        x *= f;
        return x;
    }
    template<typename T, typename... Args> inline void read(T& x, Args &...x_) {
        read(x);
        read(x_...);
        return;
    }
    inline ll read() {
        ll x;
        read(x);
        return x;
    }
    inline void read(char *s) {
        scanf("%s", s + 1);
    }
};
using namespace FastIO;

const int N = 2e5 + 10;

struct Edge {
    int to, nt, wt;
    Edge() {}
    Edge(int to, int nt, int wt) : to(to), nt(nt), wt(wt) {}
}e[N << 1];
int hd[N], cnte;
inline void AddEdge(int u, int v, int w) {
    e[++cnte] = Edge(v, hd[u], w);
    hd[u] = cnte;
}

int n;
int c[N];

inline void Input() {
    read(n);
    for(int i = 1; i <= n; i++) {
        read(c[i]);
    }
    int u, v;
    for(int i = 1; i < n; i++) {
        read(u, v);
        AddEdge(u, v, c[u] != c[v]);
        AddEdge(v, u, c[v] != c[u]);
    }
}

int dis[N];

inline void Dfs(int u, int fa) {
    for(int i = hd[u]; i; i = e[i].nt) {
        int v = e[i].to;
        if(v == fa) continue;
        dis[u] = dis[v] + e[i].wt;
        Dfs(v, u);
    }
}

inline void Work() {
    Dfs(1, 1);
    int mx = 0, dx, dy;
    for(int i = 1; i <= n; i++) {
        if(mx < dis[i]) mx = dis[i], dx = i;
    }
    memset(dis, 0, sizeof(dis));
    Dfs(dx, dx);
    mx = 0;
    for(int i = 1; i <= n; i++) {
        if(mx < dis[i]) mx = dis[i], dy = i;
    }
    printf("%d", (mx + 1) / 2);
}

int main() {
    int T = 1;
    while(T--) {
        Input();
        Work();
    }
    return 0;
}