HDU 2196 Computer（树形dp）

最新推荐文章于 2019-11-05 00:42:39 发布

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最新推荐文章于 2019-11-05 00:42:39 发布

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分类专栏：图 dp 模板树形dp 搜索

本文链接：https://blog.youkuaiyun.com/weixin_43806345/article/details/99478397

该博客探讨了如何解决一个关于带权无向树的问题，即计算每个节点到树中任意节点的最远距离。通过深度优先搜索（DFS）可以解决节点向下走的最大距离，而向上走的最大距离则需要考虑节点的父节点及其儿子节点的最大值。解决方案涉及树的遍历和动态规划，计算每个节点向上和向下走的最大路径长度。

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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

给你一个带权无向树，求每个节点到树中任意节点的最远距离

比较难。

考虑一个树，他的节点最远距离，要不向下走，要不向上走
向下走很好求，dfs一下就行
向上走的话，father可能也向上走，也可能通过别的最大的son。因此，要保存son的最大值和不走属于最大值的那个son节点的最大值，还要保存属于最大值的那个son节点的编号id【】（用于找向上最大值回溯）
答案就是向下走和向上走的最大值。

转移的时候：

一个节点向上走的最大值dp[2][now]要分类讨论，除去自己，father向下走的最大值(dp[1]father，dp[0][father]（最大值））和向上走的最大值dp[2][father]中的一个。

ps：这个题好绕口

#include <iostream>
#include <cstdio>
#include <vector>
#include <cstring>
using namespace std;
const int maxn=1e4+5;
vector<int> p[maxn],v[maxn];
int dp[3][maxn],id[maxn];

void dfs1(int x,int f){
   
   
	for(int i