题意：求节点间的最大距离

先DFS一次 记录下 每一节点的子树下的最大距离（DP[ u ] [ 0 ]）和第二大距离（DP[ u ] [ 1 ]）

用DP[ v ] [ 2 ] 表示由v的父节点来的最大距离

再取DP[ u ] [ 0 ] 与 DP[ u ][ 2 ] 的最值

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <string>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <cmath>
using namespace std;
#include <queue>
#include <stack>
#include <vector>
#include <deque>
#include <set>
#include <map>
#include <time.h>;
#define cler(arr, val)    memset(arr, val, sizeof(arr))
#define FOR(i,a,b)  for(int i=a;i<=b;i++)
#define IN   freopen ("in.txt" , "r" , stdin);
#define OUT  freopen ("out.txt" , "w" , stdout);
typedef long long  LL;
const int MAXN = 10014;
const int MAXM = 20014;
const int INF = 0x3f3f3f3f;
const int mod = 1000000007;
struct node
{int v,next;LL val;
} edge[MAXM];
int head[MAXM],tol;
LL dp[MAXN][3];
bool vis[MAXN];
void init()
{cler(head,-1);tol=0;
}
void addedge(int u,int v,LL val)
{edge[tol].v=v,edge[tol].val=val,edge[tol].next=head[u];head[u]=tol++;edge[tol].v=u,edge[tol].val=val,edge[tol].next=head[v];head[v]=tol++;
}
void dfs1(int u)
{if(vis[u]) return ;vis[u]=true;for(int i=head[u]; ~i; i=edge[i].next){int v=edge[i].v;if(!vis[v]){dfs1(v);dp[u][1]=max(dp[u][1],dp[v][0]+edge[i].val);if(dp[u][1]>dp[u][0])swap(dp[u][1],dp[u][0]);}}
}
void dfs2(int u)
{if(vis[u]) return ;vis[u]=true;for(int i=head[u]; ~i; i=edge[i].next){int v=edge[i].v,val=edge[i].val;if(!vis[v]){if(dp[u][0]>dp[v][0]+val)//dp[u][0]不是由dp[v][0]+val而来的dp[v][2]=max(dp[v][2],max(dp[u][0]+val,dp[u][2]+val));//所以从非v子树来的更长else //dp[u][0]由dp[v][0]+val而来的dp[v][2]=max(dp[v][2],max(dp[u][1]+val,dp[u][2]+val));//推断非v子树来的哪个长dfs2(v);}}
}
int main()
{
#ifndef ONLINE_JUDGEfreopen("in.txt", "r", stdin);
#endifint n;while(~scanf("%d",&n)){init();for(int i=2; i<=n; i++){int a;LL b;scanf("%d %I64d",&a ,&b );addedge(i,a,b);}cler(vis,false);cler(dp,0);dfs1(1);cler(vis,false);dfs2(1);for(int i=1;i<=n;i++)printf("%I64d\n",max(dp[i][2],dp[i][0]));}}