1.28寒假集训（玩树）-2

种树种了半天写了一题树的直径题，套模板美滋滋23333

https://vjudge.net/contest/209821#problem/A

A - 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

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

题意：给一个n，接下来n-1行每行输入c1，c2，分别是2~n编号的电脑和c1的边权c2。输出1~n电脑可以跑的最远距离。

要bfs两次比较一下↓

/*树的直径*/
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<iostream>
#include<queue>
using namespace std;
const int maxv = 1111111;

struct edge
{
	int from,to,weight,next;//该边的起始点，指向点，边权，下一个边
}edge[maxv*2];

int d[maxv],d1[maxv],d2[maxv],head[maxv];
int n,num,c1,c2;
int ans,node;//最长路径，端点值
bool vis[maxv]={false};

int bfs(int s)
{
	queue<int>q;
	fill(vis,vis+maxv,false);
	fill(d,d+maxv,0);
	q.push(s);//入队
	vis[s]=true;
	d[s]=0;
	ans=0;
	node=s;
	while(!q.empty())
	{
		int u=q.front();
		q.pop();
		for(int i=head[u];i!=-1;i=edge[i].next)
		{
			int v=edge[i].to;
			if(vis[v]==false&&d[u]+edge[i].weight>d[v])//还没访问且可松弛
			{
				vis[v]=true;
				d[v]=d[u]+edge[i].weight;
				if(ans<d[v])
				{
					ans=d[v];
					node=v;
				}
				q.push(v);
			}
		}
	}
	return node;
}

int main()
{
	while(scanf("%d",&n)!=EOF){
	fill(head,head+maxv,-1);
	num=0;
	for(int i=2;i<=n;i++)
	{
		scanf("%d %d",&c1,&c2);
		edge[num].from=c1;
		edge[num].to=i;
		edge[num].weight=c2;
		edge[num].next=head[c1];
		head[c1]=num++;
		edge[num].from=i;
		edge[num].to=c1;
		edge[num].weight=c2;
		edge[num].next=head[i];
		head[i]=num++;
	}
	int W=bfs(1);
	int Q=bfs(W);
	for(int i=1;i<=n;i++)
		d1[i]=d[i];
	bfs(Q);
	for(int i=1;i<=n;i++)
		d2[i]=d[i];
	for(int i=1;i<=n;i++)
		printf("%d\n",max(d1[i],d2[i]));
	}
	return 0;
}