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 lowest price a customer can expect from some retailers.

Input Specification:

Each input file contains one test case. For each case, The first line contains three positive numbers: N (≤10^5), 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 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. All the numbers in a line are separated by a space.

Output Specification:

For each test case, print in one line the lowest price we can expect from some retailers, accurate up to 4 decimal places, and the number of retailers that sell at the lowest price. There must be one space between the two numbers. It is guaranteed that the all the prices will not exceed 10^​10.

Sample Input:

10 1.80 1.00

3 2 3 5

1 9

1 4

1 7

0

2 6 1

1 8

0

0

0

Sample Output:

1.8362 2

#include<iostream> //类似于树的层级遍历
#include<vector>
#include<iomanip>
#include<math.h>
using namespace std;
int minlevel=9999999, num=0;
void level_traversal(vector<vector<int>> &chain, int level, int root){
if(chain[root].size()!=0){
for(int i=0; i<chain[root].size(); i++)
level_traversal(chain, level+1, chain[root][i]);
}else{
if(minlevel>level){
minlevel=level;
num=1;
}else if(minlevel==level)
num++;
}
}
int main(){
int N;
double p, r;
cin>>N>>p>>r;
vector<vector<int>> chain;
for(int i=0; i<N; i++){
int n;
cin>>n;
vector<int> temp(n,0);
for(int j=0; j<n; j++)
cin>>temp[j];
chain.push_back(temp);
}
level_traversal(chain, 0, 0);
cout<<setiosflags(ios::fixed)<<setprecision(4)<<p*pow(1+0.01*r, minlevel);
cout<<" "<<num<<endl;
return 0;
}

PAT 1106 Lowest Price in Supply Chain的更多相关文章

  1. PAT甲级——1106 Lowest Price in Supply Chain(BFS)

    本文同步发布在CSDN:https://blog.csdn.net/weixin_44385565/article/details/90444872 1106 Lowest Price in Supp ...

  2. [建树(非二叉树)] 1106. Lowest Price in Supply Chain (25)

    1106. Lowest Price in Supply Chain (25) A supply chain is a network of retailers(零售商), distributors( ...

  3. PAT 甲级 1106 Lowest Price in Supply Chain

    https://pintia.cn/problem-sets/994805342720868352/problems/994805362341822464 A supply chain is a ne ...

  4. PAT Advanced 1106 Lowest Price in 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 ...

  5. 1106. Lowest Price in Supply Chain (25)

    A supply chain is a network of retailers(零售商), distributors(经销商), and suppliers(供应商)-- everyone invo ...

  6. PAT A1106 Lowest Price in Supply Chain (25 分)——树的bfs遍历

    A supply chain is a network of retailers(零售商), distributors(经销商), and suppliers(供应商)-- everyone invo ...

  7. PAT (Advanced Level) 1106. Lowest Price in Supply Chain (25)

    简单dfs #include<cstdio> #include<cstring> #include<cmath> #include<vector> #i ...

  8. PAT甲题题解-1106. Lowest Price in Supply Chain (25)-(dfs计算树的最小层数)

    统计树的最小层数以及位于该层数上的叶子节点个数即可. 代码里建树我用了邻接链表的存储方式——链式前向星,不了解的可以参考,非常好用: http://www.cnblogs.com/chenxiwenr ...

  9. 【PAT甲级】1106 Lowest Price in Supply Chain (25分)

    题意:输入一个正整数N(<=1e5),两个小数P和R,分别表示树的结点个数和商品原价以及每下探一层会涨幅的百分比.输出叶子结点深度最小的商品价格和深度最小的叶子结点个数. trick: 测试点1 ...

随机推荐

  1. uoj#34

    模板 #include<bits/stdc++.h> #define pi acos(-1) using namespace std; ; int n, m, L, x; int r[N] ...

  2. ubuntu系统快捷键设置

    1.打开'系统设置' 2.点击键盘 3.选择快捷键,查看和修改对应的快捷键.

  3. PCB genesis连孔加除毛刺孔(槽孔与槽孔)实现方法(三)

    一.为什么 连孔加除毛刺孔 原因是 PCB板材中含有玻璃纤维, 毛刺产生位置在于2个孔相交位置,由于此处钻刀受力不均导致纤维切削不断形成毛刺 ,为了解决这个问题:在钻完2个连孔后,在相交处再钻一个孔, ...

  4. PCB genesis连孔加除毛刺孔(圆孔与槽孔)实现方法(二)

    一.为什么 连孔加除毛刺孔 原因是 PCB板材中含有玻璃纤维, 毛刺产生位置在于2个孔相交位置,由于此处钻刀受力不均导致纤维切削不断形成毛刺 ,为了解决这个问题:在钻完2个连孔后,在相交处再钻一个孔, ...

  5. 安卓中Canvas实现清屏效果

    可以在代码里面添加: paint.setXfermode(new PorterDuffXfermode(PorterDuff.Mode.CLEAR)); canvas.drawPaint(paint) ...

  6. 【POJ3255/洛谷2865】[Usaco2006 Nov]路障Roadblocks(次短路)

    题目: POJ3255 洛谷2865 分析: 这道题第一眼看上去有点懵-- 不过既然要求次短路,那估计跟最短路有点关系,所以就拿着优先队列优化的Dijkstra乱搞,搞着搞着就通了. 开两个数组:\( ...

  7. 手势识别官方教程(7)识别缩放手势用ScaleGestureDetector和SimpleOnScaleGestureListener

    1.Use Touch to Perform Scaling As discussed in Detecting Common Gestures, GestureDetector helps you ...

  8. Android 将图片网址url转化为bitmap

    public Bitmap returnBitMap(final String url){ new Thread(new Runnable() { @Override public void run( ...

  9. 升级Xcode或 MacOS编译iOS出现resource fork, Finder information, or similar detritus not allowed

    很久没有在网上留下足迹了,冒个泡吧 最近升级了Xcode,编译之前的一个项目是出现问题,问题结尾如下: resource fork, Finder information, or similar de ...

  10. android计算屏幕dp

    首先我们来了解一些基本元素: px:像素,屏幕上的点. dpi:一英寸长的直线上的像素点的数量,即像素密度.标准值是160dp. /*** 正是因为dpi值其代表的特性,所以android项目的资源文 ...