1079. Total Sales of Supply Chain (25)
A supply chain is a network of retailers(零售商), distributors(经销商), and suppliers(供应商)-- everyone involved in moving a product from supplier to customer.
Starting from one root supplier, everyone on the chain buys products from one's supplier in a price P and sell or distribute them in a price that is r% higher than P. Only the retailers will face the customers. It is assumed that each member in the supply chain has exactly one supplier except the root supplier, and there is no supply cycle.
Now given a supply chain, you are supposed to tell the total sales from all the retailers.
Input Specification:
Each input file contains one test case. For each case, the first line contains three positive numbers: N (<=105), the total number of the members in the supply chain (and hence their ID's are numbered from 0 to N-1, and the root supplier's ID is 0); P, the unit price given by the root supplier; and r, the percentage rate of price increment for each distributor or retailer. Then N lines follow, each describes a distributor or retailer in the following format:
Ki ID[1] ID[2] ... ID[Ki]
where in the i-th line, Ki is the total number of distributors or retailers who receive products from supplier i, and is then followed by the ID's of these distributors or retailers. Kj being 0 means that the j-th member is a retailer, then instead the total amount of the product will be given after Kj. All the numbers in a line are separated by a space.
Output Specification:
For each test case, print in one line the total sales we can expect from all the retailers, accurate up to 1 decimal place. It is guaranteed that the number will not exceed 1010.
Sample Input:
10 1.80 1.00
3 2 3 5
1 9
1 4
1 7
0 7
2 6 1
1 8
0 9
0 4
0 3
Sample Output:
42.4
#include<stdio.h>
#include<vector>
#include<math.h>
using namespace std; struct Node
{
int parent;
double val;
vector<int> child;
}; Node Tree[];
bool visit[]; double sum; void DFS(int root,int level,double price,double rate)
{
visit[root] = true;
if(Tree[root].child.empty())
{
sum += Tree[root].val * price*pow((+rate)/,level);
}
else
{
for(int i = ;i < Tree[root].child.size();i++)
{
if(visit[Tree[root].child[i]] == false)
DFS(Tree[root].child[i],level+,price,rate);
}
}
} int main()
{
int ID,pro,count,j,num,i;
double price,rate;
scanf("%d%lf%lf",&num,&price,&rate);
for( i = ; i < num ;i++)
{
Tree[i].child.clear();
Tree[i].parent = -;
visit[i]= false;
}
for(i = ;i < num ;i ++)
{
scanf("%d",&count);
if(count == )
{
scanf("%d",&pro);
Tree[i].val = pro;
}
else
{
for(j = ;j < count ;j++)
{
scanf("%d",&ID);
Tree[i].child.push_back(ID);
Tree[ID].parent = i;
}
}
} int root = ;
while(Tree[root].parent != -)
++root;
int level = ;
sum = ;
DFS(root,level,price,rate);
printf("%0.1lf\n",sum); return ;
}
1079. Total Sales of Supply Chain (25)的更多相关文章
- 1079. Total Sales of Supply Chain (25)【树+搜索】——PAT (Advanced Level) Practise
题目信息 1079. Total Sales of Supply Chain (25) 时间限制250 ms 内存限制65536 kB 代码长度限制16000 B A supply chain is ...
- PAT 甲级 1079 Total Sales of Supply Chain (25 分)(简单,不建树,bfs即可)
1079 Total Sales of Supply Chain (25 分) A supply chain is a network of retailers(零售商), distributor ...
- 1079. Total Sales of Supply Chain (25)-求数的层次和叶子节点
和下面是同类型的题目,只不过问的不一样罢了: 1090. Highest Price in Supply Chain (25)-dfs求层数 1106. Lowest Price in Supply ...
- 1079. Total Sales of Supply Chain (25) -记录层的BFS改进
题目如下: A supply chain is a network of retailers(零售商), distributors(经销商), and suppliers(供应商)-- everyon ...
- PAT Advanced 1079 Total Sales of Supply Chain (25) [DFS,BFS,树的遍历]
题目 A supply chain is a network of retailers(零售商), distributors(经销商), and suppliers(供应商)– everyone in ...
- PAT (Advanced Level) 1079. Total Sales of Supply Chain (25)
树的遍历. #include<cstdio> #include<cstring> #include<cmath> #include<vector> #i ...
- 【PAT甲级】1079 Total Sales of Supply Chain (25 分)
题意: 输入一个正整数N(<=1e5),表示共有N个结点,接着输入两个浮点数分别表示商品的进货价和每经过一层会增加的价格百分比.接着输入N行每行包括一个非负整数X,如果X为0则表明该结点为叶子结 ...
- pat1079. Total Sales of Supply Chain (25)
1079. Total Sales of Supply Chain (25) 时间限制 250 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHE ...
- PAT 1079 Total Sales of Supply Chain[比较]
1079 Total Sales of Supply Chain(25 分) A supply chain is a network of retailers(零售商), distributors(经 ...
随机推荐
- Java设计模式12:常用设计模式之外观模式(结构型模式)
1. Java之外观模式(Facade Pattern) (1)概述: 现代的软件系统都是比较复杂的,设计师处理复杂系统的一个常见方法便是将其"分而治之",把一个系统划 ...
- hadoop学习记录(三)HBase基本概念
这一次开始学习HBase数据库. 我用的是VMWare + ubuntu16.04 +Hbase1.1.5 +hadoop2.6.0的组合. 经过亲自安装验证,版本间没有发生冲突,可以进行学习和开发. ...
- iOS “获取验证码”按钮的倒计时功能
iOS 的倒计时有多种实现细节,Cocoa Touch 为我们提供了 NSTimer 类和 GCD 的dispatch_source_set_timer方法去更加方便的使用计时器.我们也可以很容易的的 ...
- 给jdk写注释系列之jdk1.6容器(8)-TreeSet&NavigableMap&NavigableSet源码解析
TreeSet是一个有序的Set集合. 既然是有序,那么它是靠什么来维持顺序的呢,回忆一下TreeMap中是怎么比较两个key大小的,是通过一个比较器Comparator对不对,不过遗憾的是,今天仍然 ...
- 每天一道LeetCode--342. Power of Four
Given an integer (signed 32 bits), write a function to check whether it is a power of 4. Example:Giv ...
- [转]SQL中使用WITH AS提高性能-使用公用表表达式(CTE)简化嵌套SQL
一.WITH AS的含义 WITH AS短语,也叫做子查询部分(subquery factoring),可以让你做很多事情,定义一个SQL片断,该SQL片断会被整个SQL语句所用到.有的时候, ...
- 一个简单的Makefile的编写【用自己的话,解释清楚这些】
用自己的话,解释清楚这些~ Makefile是程序员编写出来指导编译器编译程序源码为目标文件(可执行文件,或链接库) 这里只写一个简单的Makefile 作为例子 其需求如下: frank@ubunt ...
- Linux 最常用命令小结
1. 文件共享 1).将windows 系统下的文件夹共享到linux的方法: 安装filezilla,设置连接linux 服务器.将文件上传. 2).mRemote 机器连接管理 2. 文件管理命令 ...
- Swift属性
属性的存储 属性的主要作用是存储数据,可以常量属性和变量属 性: struct FixedLengthRange { var firstValue: Int let length: Int } var ...
- OC4_实例变量的作用域
// // Dog.h // OC4_实例变量的作用域 // // Created by zhangxueming on 15/6/16. // Copyright (c) 2015年 zhangxu ...