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给定一颗 $N$ 个节点的树，结点从 $1$ 到 $N$ 编号。

现在要计算对于每个结点 $i$ ，到它最远的结点是多少距离。

思路

显而易见的，树上最远的点对是树的直径的两个端点。而学过小学数学的人都知道，从 $i$ 点出发，经过最远的一条链的一部分，最后到达这条最远链的两个端点之一一定是最长的。

代码

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/**
 * @file P2386C.cpp
 * @author Zhoumouren
 * @brief 
 * @version 0.1
 * @date 2024-08-22
 * 
 * @copyright Copyright (c) 2024

As All We know ,,, 
the maximum distance form a node i is the two side of 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 = 1e4 + 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;
void AddEdge(int u, int v, int w) {
    e[++cnte] = Edge(v, hd[u], w);
    hd[u] = cnte;
}

int n;

inline void Input() {
    read(n);
    int u, w;
    for(int i = 1; i < n; i++) {
        read(u, w);
        AddEdge(u, i + 1, w);
        AddEdge(i + 1, u, w);
    }
}

int dis[N][2];
int cur;

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[v][cur] = dis[u][cur] + e[i].wt;
        Dfs(v, u);
    }
}

inline void Work() {
    cur = 0;
    Dfs(1, 1);
    int mx = 0, du, dv;
    for(int i = 1; i <= n; i++) {
        if(mx < dis[i][cur]) {
            mx = dis[i][cur];
            du = i;
        }
        dis[i][cur] = 0;
    }
    Dfs(du, du);
    mx = 0;    
    for(int i = 1; i <= n; i++) {
        if(mx < dis[i][cur]) {
            mx = dis[i][cur];
            dv = i;
        }
    }
    cur = 1;
    Dfs(dv, dv);
    for(int i = 1; i <= n; i++) {
        printf("%d\n", max(dis[i][0], dis[i][1]));
    }
}

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