2013
12-30

# Computer

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 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.

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.

5
1 1
2 1
3 1
1 1

3
2
3
4
4

/*
HDU 2196
G++ 46ms  916K

*/

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<string.h>
using namespace std;
const int MAXN=10010;
struct Node
{
int to;
int next;
int len;
}edge[MAXN*2];//因为存无向边，所以需要2倍
int tol;
int maxn[MAXN];//该节点往下到叶子的最大距离
int smaxn[MAXN];//次大距离
int maxid[MAXN];//最大距离对应的序号
int smaxid[MAXN];//次大的序号

void init()
{
tol=0;
}

{
edge[tol].to=b;
edge[tol].len=len;
edge[tol].to=a;
edge[tol].len=len;
}

//求结点v往下到叶子结点的最大距离
//p是v的父亲结点
void dfs1(int u,int p)
{
maxn[u]=0;
smaxn[u]=0;
{
int v=edge[i].to;
if(v==p)continue;//不能往上找父亲结点
dfs1(v,u);
if(smaxn[u]<maxn[v]+edge[i].len)
{
smaxn[u]=maxn[v]+edge[i].len;
smaxid[u]=v;
if(smaxn[u]>maxn[u])
{
swap(smaxn[u],maxn[u]);
swap(smaxid[u],maxid[u]);
}
}
}
}
//p是u的父亲结点，len是p到u的长度
void dfs2(int u,int p)
{
{
int v=edge[i].to;
if(v==p)continue;
if(v==maxid[u])
{
if(edge[i].len+smaxn[u]>smaxn[v])
{

smaxn[v]=edge[i].len+smaxn[u];
smaxid[v]=u;
if(smaxn[v]>maxn[v])
{
swap(smaxn[v],maxn[v]);
swap(smaxid[v],maxid[v]);
}
}
}
else
{
if(edge[i].len+maxn[u]>smaxn[v])
{
smaxn[v]=edge[i].len+maxn[u];
smaxid[v]=u;
if(smaxn[v]>maxn[v])
{
swap(smaxn[v],maxn[v]);
swap(maxid[v],smaxid[v]);
}
}
}
dfs2(v,u);
}
}
int main()
{
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
int n;
int v,len;
while(scanf("%d",&n)!=EOF)
{
init();
for(int i=2;i<=n;i++)
{
scanf("%d%d",&v,&len);
}
dfs1(1,-1);

dfs2(1,-1);
for(int i=1;i<=n;i++)
printf("%d\n",maxn[i]);
}
return 0;
}

1. 我还有个问题想请教一下，就是感觉对于新手来说，递归理解起来有些困难，不知有没有什么好的方法或者什么好的建议？