1079 Total Sales of Supply Chain
来源:互联网 发布:锦衣卫知乎 编辑:程序博客网 时间:2024/06/06 01:03
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
解题思路:因为同类型的题目写过了,所以写起来很快,一遍过。
#include<iostream>#include<stdio.h>#include<vector>#include<math.h>using namespace std;int main(){ int n; double price; double rate; for (; scanf("%d%lf%lf", &n, &price, &rate) != EOF;){ int *pro_amount = new int[n]; vector<int> *nodes = new vector<int>[n]; for (int i = 0; i < n; i++){ int size; scanf("%d", &size); if (size){ nodes[i] = vector<int>(size); for (int j = 0; j < size;j++){ scanf("%d", &nodes[i][j]); } pro_amount[i] = 0; } else{ nodes[i] = vector<int>(size); scanf("%d", &pro_amount[i]); } } int level = 0; double sum = 0.0; vector<int>vec; vec.push_back(0); while (vec.size()){ vector<int>temp; for (int i = 0; i < vec.size(); i++){ //是零售商 if (pro_amount[vec[i]]){ sum += price*pow(rate / 100.0 + 1, level)*pro_amount[vec[i]]; } else{ for (int j = 0; j < nodes[vec[i]].size(); j++){ temp.push_back(nodes[vec[i]][j]); } } } level++; vec = vector<int>(temp); } printf("%.1lf\n", sum); } return 0;}
- pat 1079 Total Sales of Supply Chain
- 1079 Total Sales of Supply Chain
- Pat(Advanced Level)Practice--1079(Total Sales of Supply Chain)
- PAT 1079-Total Sales of Supply Chain (25)
- 1079. Total Sales of Supply Chain (25)
- 1079. Total Sales of Supply Chain (25)
- 1079. Total Sales of Supply Chain (25)
- 1079.Total Sales of Supply Chain
- 1079. Total Sales of Supply Chain (25)
- PAT1079 Total Sales of Supply Chain
- pat1079 Total Sales of Supply Chain
- 1079. Total Sales of Supply Chain (25)
- 1079. Total Sales of Supply Chain (25)
- 1079. Total Sales of Supply Chain (25)
- 1079. Total Sales of Supply Chain (25)
- 1079. Total Sales of Supply Chain (25)
- 1079. Total Sales of Supply Chain
- 1079. Total Sales of Supply Chain (25)
- 二叉树的性质
- 【cdq分治&NTT】BNUOJ51279组队活动
- Android Studio:Button控件中英文全部显示为大写的解决方法
- SQLite应用程序启动报错:Object reference not set to an instance of an object.
- NUBT 1222 English Game 【字典树 + dp】
- 1079 Total Sales of Supply Chain
- linux sed命令详解
- Linux如何查找大文件和大目录
- linux分配虚拟内存(swap)
- uploader秒传图片到服务器完整版
- 浙江省人民政府关于建立新型农村合作医疗制度的实施意见
- leetcode笔记:Range Sum Query 2D - Immutable
- 启动报Cannot get connection for url jdbc xxxx listener could not hand off client co
- uva 11054 Gergovia的酒交易 等价转化/贪心