Farmer John's family pitches in with the chores during milking, doing all the chores as quickly as possible. At FJ's house, some chores cannot be started until others have been completed, e.g., it is impossible to wash the cows until they are in the stalls.

Farmer John has a list of N (3 <= N <= 10,000) chores that must be completed. Each chore requires an integer time (1 <= length of time <= 100) to complete and there may be other chores that must be completed before this chore is started. We will call these prerequisite chores. At least one chore has no prerequisite: the very first one, number 1. Farmer John's list of chores is nicely ordered, and chore K (K > 1) can have only chores 1,.K-1 as prerequisites. Write a program that reads a list of chores from 1 to N with associated times and all perquisite chores. Now calculate the shortest time it will take to complete all N chores. Of course, chores that do not depend on each other can be performed simultaneously.

Input

* Line 1: One integer, N

* Lines 2..N+1: N lines, each with several space-separated integers. Line 2 contains chore 1; line 3 contains chore 2, and so on. Each line contains the length of time to complete the chore, the number of the prerequisites, Pi, (0 <= Pi <= 100), and the Pi prerequisites (range 1..N, of course).

Output

A single line with an integer which is the least amount of time required to perform all the chores. 

Sample Input

7
5 0
1 1 1
3 1 2
6 1 1
1 2 2 4
8 2 2 4
4 3 3 5 6

Sample Output

23

Hint

[Here is one task schedule:

Chore 1 starts at time 0, ends at time 5.

Chore 2 starts at time 5, ends at time 6.

Chore 3 starts at time 6, ends at time 9.

Chore 4 starts at time 5, ends at time 11.

Chore 5 starts at time 11, ends at time 12.

Chore 6 starts at time 11, ends at time 19.

Chore 7 starts at time 19, ends at time 23.

]
题解:树形DP入门题。从根节点往下依次更新出每一个节点的最短时间,则该最短时间的最大值即为:完成家务的最短时间。
参考代码为:
#include <iostream>
#include <cstring>
using namespace std;
const int maxn=10005;
int c[maxn],n[maxn],dp[maxn]; int main()
{
ios::sync_with_stdio(false);
cin.tie(0);
int N,temp,sum=-maxn;
memset(dp,0,sizeof dp);
cin>>N;
for(int i=1;i<=N;i++)
{
cin>>c[i]>>n[i];
if(i==1) dp[i]=c[i];
else
{
int max=-maxn;
if(n[i]==0) dp[i]=c[i];
else
{
for(int j=0;j<n[i];j++)
{
cin>>temp;
if(dp[temp]>max) max=dp[temp];
}
dp[i]=max+c[i];
}
}
if(dp[i]>sum) sum=dp[i];
}
cout<<sum<<endl;
return 0;
} /*
7
5 0
1 1 1
3 1 2
6 1 1
1 2 2 4
8 2 2 4
4 3 3 5 6
*/

  

POJ 1949 Chores的更多相关文章

  1. POJ 1949 Chores (很难想到的dp)

    传送门: http://poj.org/problem?id=1949 Chores Time Limit: 3000MS   Memory Limit: 30000K Total Submissio ...

  2. poj 1949 Chores 最长路

    题目链接 求出最长路..... #include <iostream> #include <vector> #include <cstdio> #include & ...

  3. POJ 1949 Chores(DAG上的最长路 , DP)

    题意: 给定n项任务, 每项任务的完成用时t和完成每项任务前需要的k项任务, 求把所有任务完成的最短时间,有当前时间多项任务都可完成, 那么可以同时进行. 分析: 这题关键就是每项任务都会有先决条件, ...

  4. POJ 1949 DP?

    题意: 有n个家务,第i个家务需要一定时间来完成,并且第i个任务必须在它 "前面的" 某些任务完成之后才能开始. 给你任务信息,问你最短需要多少时间来完成任务. 输入: 第一行n个 ...

  5. poj 动态规划题目列表及总结

    此文转载别人,希望自己能够做完这些题目! 1.POJ动态规划题目列表 容易:1018, 1050, 1083, 1088, 1125, 1143, 1157, 1163, 1178, 1179, 11 ...

  6. poj动态规划列表

    [1]POJ 动态规划题目列表 容易: 1018, 1050, 1083, 1088, 1125, 1143, 1157, 1163, 1178, 1179, 1189, 1208, 1276, 13 ...

  7. 【转载】图论 500题——主要为hdu/poj/zoj

    转自——http://blog.csdn.net/qwe20060514/article/details/8112550 =============================以下是最小生成树+并 ...

  8. POJ 动态规划题目列表

    ]POJ 动态规划题目列表 容易: 1018, 1050, 1083, 1088, 1125, 1143, 1157, 1163, 1178, 1179, 1189, 1208, 1276, 1322 ...

  9. poj 动态规划的主题列表和总结

    此文转载别人,希望自己可以做完这些题目. 1.POJ动态规划题目列表 easy:1018, 1050, 1083, 1088, 1125, 1143, 1157, 1163, 1178, 1179, ...

随机推荐

  1. Linux下mysql的主从复制配置

    1.准备两台数据库环境,或者单台多实例环境,能正常启动和登录. 数据库的安装和多实例配置请参考https://www.cnblogs.com/qiuhom-1874/p/9757061.html. 2 ...

  2. 013.Kubernetes二进制部署worker节点Nginx实现高可用

    一 Nginx代理实现kube-apiserver高可用 1.1 Nginx实现高可用 基于 nginx 代理的 kube-apiserver 高可用方案. 控制节点的 kube-controller ...

  3. SqlServer2005 查询 第一讲 计算列

    数据库查询[最重要且在数据库中难度最大] 声明一下我这里用的数据库样例是郝斌老师的(scott库) 我尽最大努力把复杂的问题简单化,方便理解,希望我写的东西能够帮助到你们 有些复杂的东西我我用自己的方 ...

  4. Mirantis 收购 Docker | 云原生生态周报 Vol. 28

    作者 | 禅鸣.进超.心水.心贵 业界要闻 Docker 将 Docker Enterprise 卖给 Mirantis Mirantis 是一家扎根于 OpenStack 的云公司,最近专注于 Ku ...

  5. Centos上通过shell脚本备份数据库

    #!/bin/bash ds=`` list=`date +%Y`/`date +%m` dname="callme" eval "mkdir -p $list" ...

  6. Java每日一面(Part1:计算机网络)[19/11/25]

    作者:晨钟暮鼓c个人微信公众号:程序猿的月光宝盒 1. HTTP相关[2] 1.1Get请求和Post请求的区别 从三个层面来回答: 1.1.1 从HTTP报文层面: ​ Get请求将请求信息放在UR ...

  7. es5设置属性不能修改

    /*es5*/ { var Person ={ name:'es5', age:19 } Object.defineProperty(Person,'sex',{ writable:false, va ...

  8. 预分配——fallocate的前世今生

    最近比较懒,还是加班写点东西吧,不然过段时间又把这些整理的东西弄丢了. 写什么呢?写一些跟工作相关的吧!因为笔者从事多媒体录像相关的开发工作,因此常常涉及到优化写卡策略.提升写卡性能相关的方面的事情. ...

  9. .Net Core 3.0 使用 Serilog 把日志记录到 SqlServer

    Serilog简介 Serilog是.net中的诊断日志库,可以在所有的.net平台上面运行.Serilog支持结构化日志记录,对复杂.分布式.异步应用程序的支持非常出色.Serilog可以通过插件的 ...

  10. 详解在Linux系统中安装Tomcat

    本文以在CentOS 7.6中安装Tomcat8.5为例进行安装,其他系统和版本都是大同小异的. 安装JDK 安装Tomcat之前,需要先安装JDK,可以参看之前的文章详解在Linux系统中安装JDK ...