Computer HDU

A school bought the first computer some time ago(so this computer's id is 1). During the recent years the school bought N-1 new computers. Each new computer was connected to one of settled earlier. Managers of school are anxious about slow functioning of the net and want to know the maximum distance Si for which i-th computer needs to send signal (i.e. length of cable to the most distant computer). You need to provide this information.

Hint: the example input is corresponding to this graph. And from the graph, you can see that the computer 4 is farthest one from 1, so S1 = 3. Computer 4 and 5 are the farthest ones from 2, so S2 = 2. Computer 5 is the farthest one from 3, so S3 = 3. we also get S4 = 4, S5 = 4.InputInput file contains multiple test cases.In each case there is natural number N (N<=10000) in the first line, followed by (N-1) lines with descriptions of computers. i-th line contains two natural numbers - number of computer, to which i-th computer is connected and length of cable used for connection. Total length of cable does not exceed 10^9. Numbers in lines of input are separated by a space.OutputFor each case output N lines. i-th line must contain number Si for i-th computer (1<=i<=N).Sample Input

5
1 1
2 1
3 1
1 1

Sample Output

3
2
3
4
4
题解：直接对每个点进行搜索的话肯定会超时，这里涉及到树的直径问题，记直径的一段为a，另一端记为b。那么树中任意一点所能走到的最长距离就是max(到a的距离，到b的距离).剩下的就是在求直径的过程中求每个点到根(既端点)的距离

#include<vector>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;

const int maxn=1e5+5;

struct node{
    int to,next,va;
}e[maxn];

int n,tot;
int head[maxn],d1[maxn],d2[maxn];

void Inite(){
    tot=0;
    memset(head,-1,sizeof(head));
}

void addedge(int u,int v,int w){
    e[tot].to=v;
    e[tot].va=w;
    e[tot].next=head[u];
    head[u]=tot++;
}

void DFS(int pa,int u,int d[]){
    for(int i=head[u];i!=-1;i=e[i].next){
        int v=e[i].to;
        if(pa==v) continue;
        d[v]=d[u]+e[i].va;
        DFS(u,v,d);
    }
}

int main()
{   while(~scanf("%d",&n)){
        Inite();
        for(int i=2;i<=n;i++){
            int u,va;
            scanf("%d%d",&u,&va);
            addedge(i,u,va);
            addedge(u,i,va);
        }
        memset(d1,0,sizeof(d1));
        DFS(0,1,d1);
        
        int temp=0;
        for(int i=1;i<=n;i++) if(d1[temp]<d1[i]) temp=i;
        
        memset(d1,0,sizeof(d1));
        DFS(0,temp,d1);
          
        temp=0;
        for(int i=1;i<=n;i++) if(d1[temp]<d1[i]) temp=i;
        
        memset(d2,0,sizeof(d2));
        DFS(0,temp,d2);
        
        for(int i=1;i<=n;i++) cout<<max(d1[i],d2[i])<<endl;
    }
}