Computer

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 5925    Accepted Submission(s): 2979

Problem Description

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.

 

 

Input

Input 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.

 

 

Output

For 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

 

 

Author

scnu

 

 

Recommend

lcy

 

#include
      
       #include
       
        #include
        
         #include
         
          #include
          
           #define N 10010using namespace std;struct node{    int to,len;//下一个节点，长度    node (int x,int y)    {        to=x;        len=y;    }};int n;/*第一个dfs是先搜然后再转移状态的，用结点下方的数据更新第二个dfs是先状态转移然后再搜的，这是从结点上方进行数据更新，刚好满足题意：结点左右两个“子树”*/vector
           
             adj[N*2];int maxn[N];//第i个点的最大权值int smaxn[N];//第i个点的第二大的权值int maxi[N];//最大权值的点int smaxi[N];//第二大权值的点void dfs1(int u,int p)//p是u的父节点{ maxn[u]=0; smaxn[u]=0; for(int i=0;i
            
             smaxn[u])//先和第二长的距离比较，这样能记录下第二长的距离，这个数据能为下一个dfs提供判断 { smaxn[u]=maxn[v]+adj[u][i].len; smaxi[u]=v; if(maxn[u]
             
              smaxn[v]) { smaxn[v]=adj[u][i].len+smaxn[u]; smaxi[v]=u; if(maxn[v]
              
               smaxn[v]) { smaxn[v]=adj[u][i].len+maxn[u]; smaxi[v]=u; if(maxn[v]