HDOJ 2196 Computer

最新推荐文章于 2019-08-15 15:54:41 发布

原创最新推荐文章于 2019-08-15 15:54:41 发布 · 1.1k 阅读

0 ·

CC 4.0 BY-SA版权

DP 专栏收录该内容

150 篇文章

订阅专栏

本文介绍了一种使用树形动态规划算法解决学校计算机网络中寻找每台计算机到网络中最远计算机的距离的问题。通过两次深度优先搜索实现，首先确定每个节点的最长和次长路径，然后计算每个节点作为根时的最远距离。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

经典的树形DP。。。

Computer

Time Limit: 1000MS

Memory Limit: 32768KB

64bit IO Format: %I64d & %I64u

[Submit] [Go Back] [Status]

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

Sample Output

Source

[Submit] [Go Back] [Status]

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int maxn=22000;
const int INF=0x3f3f3f3f;

int Adj[maxn],Size;

struct E
{
	int to,next,len;
}Edge[maxn];

void init()
{
	Size=0;
	memset(Adj,-1,sizeof(Adj));
}

void Add_Edge(int u,int v,int w)
{
	Edge[Size].to=v;
	Edge[Size].len=w;
	Edge[Size].next=Adj[u];
	Adj[u]=Size++;
}

int n,dp[maxn][3];

void dfs1(int f,int u)
{
    int biggest=0,bigger=0;
    for(int i=Adj[u];~i;i=Edge[i].next)
    {
        int v=Edge[i].to;
        int w=Edge[i].len;
        if(v==f) continue;
        dfs1(u,v);
        int temp=dp[v][0]+w;
        if(temp>=biggest)
        {
            bigger=biggest;
            biggest=temp;
        }
        else if(temp>bigger)
        {
            bigger=temp;
        }
    }
    dp[u][0]=biggest;
    dp[u][1]=bigger;
}

void dfs2(int f,int u)
{
    for(int i=Adj[u];~i;i=Edge[i].next)
    {
        int v=Edge[i].to;
        int w=Edge[i].len;
        if(v==f) continue;
        dp[v][2]=max(dp[u][2],(dp[v][0]+w==dp[u][0])?dp[u][1]:dp[u][0]) + w ;
        dfs2(u,v);
    }
}

int main()
{
while(scanf("%d",&n)!=EOF)
{
	init();
	for(int i=2;i<=n;i++)
	{
		int a,b;
		scanf("%d%d",&a,&b);
		Add_Edge(i,a,b);
		Add_Edge(a,i,b);
	}
	memset(dp,0,sizeof(dp));
    dfs1(-1,1);
    dfs2(-1,1);
	for(int i=1;i<=n;i++)
    {
        printf("%d\n",max(dp[i][0],dp[i][2]));
    }
}
	return 0;
}