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
1 1
2 1
3 1
1 1
Sample Output
3
2
3
4
4
2
3
4
4
Author
scnu
Recommend
lcy
#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<string.h>
#include<vector>
#define N 10010
using 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<node> adj[N*];
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]=;
smaxn[u]=;
for(int i=;i<adj[u].size();i++)
{
int v=adj[u][i].to;
if(v==p) continue;//和父节点一样
dfs1(v,u);
if(maxn[v]+adj[u][i].len>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[u])//更新之后第二长的路径，如果比原来最长的路径还长就替换，并且原来的最长的就变成第二长得了
{
swap(maxn[u],smaxn[u]);
swap(maxi[u],smaxi[u]);
}
}
}
} //从父节点更新过来的
void dfs2(int u,int p)
{
for(int i=;i<adj[u].size();i++)
{
int v=adj[u][i].to;
if(v==p) continue;//和父节点一样
if(v==maxi[u])
//这个地方如果下一个结点刚巧是就是dfs1中搜出来的最大值的节点的话
//那么这一条路就不能用最大值去算，因为这样就算上上一条路的了
{
if(adj[u][i].len+smaxn[u]>smaxn[v])
{
smaxn[v]=adj[u][i].len+smaxn[u];
smaxi[v]=u;
if(maxn[v]<smaxn[v])
{
swap(maxn[v],smaxn[v]);
swap(maxi[v],smaxi[v]);
}
}
}
else
{
if(adj[u][i].len+maxn[u]>smaxn[v])
{
smaxn[v]=adj[u][i].len+maxn[u];
smaxi[v]=u;
if(maxn[v]<smaxn[v])
{
swap(maxn[v],smaxn[v]);
swap(maxi[v],smaxi[v]);
}
}
}
dfs2(v,u);
}
}
int main()
{
//freopen("in.txt","r",stdin);
while(scanf("%d",&n)!=EOF)
{
for(int i=;i<=n;i++)
adj[i].clear();
for(int i=;i<=n;i++)
{
int a,b;
scanf("%d%d",&a,&b);
adj[i].push_back(node(a,b));
adj[a].push_back(node(i,b));
}
dfs1(,-);//从子节点更新
dfs2(,-);//从父节点更新
for(int i=;i<=n;i++)
printf("%d\n",maxn[i]);
}
return ;
}