题目大意:给N个点,求每个点的与其他点距离最大值

很经典的树形dp...很久前就想写来着...看了陈老师的code才会的...mx[x][0], mx[x][1]分别表示x点子树里最长的2个距离, dfs一遍得到. mx[x][2]表示从x的父亲到x的最长路径长度, 也是dfs一遍得到(具体看代码)。最后答案就是max(mx[x][0], mx[x][2]). 时间复杂度O(N)

--------------------------------------------------------------------------------------------

#include<cstdio>
#include<cstring>
#include<algorithm>
 
using namespace std;
 
const int maxn = 10009;
 
int N, mx[maxn][3];
 
struct edge {
int to, w;
edge* next;
} E[maxn << 1], *pt = E, *head[maxn];
 
void AddEdge(int u, int v, int w) {
pt->to = v; pt->w = w; pt->next = head[u]; head[u] = pt++;
}
 
void Init() {
scanf("%d", &N);
for(int i = 1; i < N; i++) {
int t, w; scanf("%d%d", &t, &w); t--;
AddEdge(i, t, w);
AddEdge(t, i, w);
}
}
 
inline void upd(int x, int t) {
if(mx[x][1] < t)
mx[x][1] = t;
if(mx[x][0] < mx[x][1])
swap(mx[x][0], mx[x][1]);
}
 
void DFS0(int x, int fa = -1) {
mx[x][0] = mx[x][1] = 0;
for(edge* e = head[x]; e; e = e->next) if(e->to != fa) {
DFS0(e->to, x);
upd(x, mx[e->to][0] + e->w);
}
}
 
void DFS1(int x, int fa = - 1) {
for(edge* e = head[x]; e; e = e->next) if(e->to != fa) {
mx[e->to][2] = mx[x][2] + e->w;
if(mx[e->to][0] + e->w == mx[x][0])
mx[e->to][2] = max(mx[e->to][2], mx[x][1] + e->w);
else
mx[e->to][2] = max(mx[e->to][2], mx[x][0] + e->w);
DFS1(e->to, x);
}
}
 
int main() {
Init();
DFS0(0);
DFS1(mx[0][2] = 0);
for(int i = 0; i < N; i++)
printf("%d\n", max(mx[i][2], mx[i][0]));
return 0;
}

--------------------------------------------------------------------------------------------

149. Computer Network

time limit per test: 0.25 sec.
memory limit per test: 4096 KB
input: standard input
output: standard output

A school bought the first computer some time ago. 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 for each computer number Si - maximum distance, 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.
Input
There is natural number N (N<=10000) in the first line of input, 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
Write N lines in output file. i-th line must contain number Si for i-th computer (1<=i<=N).
Sample test(s)
Input


1 1 
1 2
Output



3

Author: Andrew V. Lazarev, Michael R. Mirzayanov
Resource: Saratov Subregional School Team Contest, 2002
Date: Fall, 2002

SGU 149. Computer Network( 树形dp )的更多相关文章

  1. SGU 149 Computer Network 树DP/求每个节点最远端长度

    一个比较经典的题型,两次DFS求树上每个点的最远端距离. 参考这里:http://hi.baidu.com/oi_pkqs90/item/914e951c41e7d0ccbf904252 dp[i][ ...

  2. SGU 149. Computer Network

    时间限制:0.25s 空间限制:4M: 题意: 给出一颗n(n<=10000)个节点的树,和n-1条边的长度.求出这棵树每个节点到最远节点的距离: Solution: 对于一个节点,我们可以用D ...

  3. HDU2196 Computer(树形DP)

    和LightOJ1257一样,之前我用了树分治写了.其实原来这题是道经典的树形DP,感觉这个DP不简单.. dp[0][u]表示以u为根的子树中的结点与u的最远距离 dp[1][u]表示以u为根的子树 ...

  4. poj3417 Network 树形Dp+LCA

    题意:给定一棵n个节点的树,然后在给定m条边,去掉m条边中的一条和原树中的一条边,使得树至少分为两部分,问有多少种方案. 神题,一点也想不到做法, 首先要分析出加入一条边之后会形成环,形成环的话,如果 ...

  5. Uva LA 3902 - Network 树形DP 难度: 0

    题目 https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&Itemid=8&page=show_pr ...

  6. HDU 2136:Computer(树形DP)

    http://acm.split.hdu.edu.cn/showproblem.php?pid=2196 Computer     Description A school bought the fi ...

  7. hdu2196 Computer【树形DP】【换根法】

    Computer Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Su ...

  8. 题解报告:hdu 2196 Computer(树形dp)

    Problem Description A school bought the first computer some time ago(so this computer's id is 1). Du ...

  9. Computer (树形DP)

    A school bought the first computer some time ago(so this computer's id is 1). During the recent year ...

随机推荐

  1. Visual Studio使用技巧

    编程部分: 1.TODO:书签 打开之后返回上次工作的位置.让我们不再用脑子去记忆,去回顾刚刚工作到的部分. 操作非常easy.例如以下代码所看到的: public DataTable SelectB ...

  2. fragment详解(官方文档)

    原作者为: 苍山.感谢他分享的内容,现在分享出来给eoeAndroid的各位同胞. 详情参考http://www.eoeandroid.com/thread-71642-1-1.html和http:/ ...

  3. ssh框架的搭建

    SSH 为 struts+spring+hibernate的一个集成框架,是目前较流行的一种Web应用程序开源框架. 集成SSH框架的系统从职责上分为四层:表示层.业务逻辑层.数据持久层和域模块层,以 ...

  4. C++_内部类

    C++ 内部类和外部类之间的相互调用

  5. unix网络编程笔记

    TCP连接状态转换图:

  6. 加密传输SSL协议3_非对称加密

    困死了,这里对非对称加密体系开个头,具体的实验明天写 非对称加密体系 为了解决对称加密中密钥的传输的问题,一些天才的数学家就提出了非对称式的加密体系,也称为公钥加密体系. 加密和解密的密钥是不同的.一 ...

  7. python打包成.exe工具py2exe0-----No such file or directory错误

    转自:http://justcoding.iteye.com/blog/900993 一.简介 py2exe是一个将python脚本转换成windows上的可独立执行的可执行程序(*.exe)的工具, ...

  8. 动态include与静态include的区别

    搬一下以前写的 个人总结: 动态INCLUDE 用jsp:include动作实现 <jsp:include page="included.jsp" flush="t ...

  9. sheelエラー、オブジェクトを解析中にエラーが発生しました。

  10. jmeter实例演示

    Jmeter 是比较轻便的性能测试工具,下面根据一个实例演示下jmeter的常见用法 一.前期准备,在使用之前,先分析测试需求,比如:需不要登录?需不需要监视服务器性能?多线程还是多循环?需不需根据流 ...