pat 1079 Total Sales of Supply Chain
来源:互联网 发布:如果国民党赢了 知乎 编辑:程序博客网 时间:2024/06/14 11:14
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 1.003 2 3 51 91 41 70 72 6 11 80 90 40 3Sample Output:
42.4
#include <iostream>#include <cstdio>#include <vector>#include <map>#include <utility>using namespace std;vector<vector<int> > vec(100010);map<int, long long> leaf;int n;double p, r;double sum = 0.0;void dfs(int index, double price) {if(leaf.count(index) > 0) {sum += price * leaf[index];return;}for (int i = 0; i < vec[index].size(); ++i){dfs(vec[index][i], price*(1+r/100.0));}}int main() {scanf("%d %lf %lf", &n, &p, &r);for (int i = 0; i < n; ++i) { int tmp; scanf("%d", &tmp); if(tmp == 0) { long long x; scanf("%lld", &x); leaf[i]=x; } for(int j = 0; j < tmp; j++) { int x; scanf("%d", &x); vec[i].push_back(x); } } dfs(0, p); printf("%.1lf\n", sum); return 0;}
要用 double 类型 和 long long 类型, 如果用 float 和int ,好几个测试点会通不过。
- pat 1079 Total Sales of Supply Chain
- 【PAT】1079. Total Sales of Supply Chain
- PAT--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
- PAT 1079. Total Sales of Supply Chain (25)
- PAT (Advanced) 1079. Total Sales of Supply Chain (25)
- PAT Advanced Level 1079. Total Sales of Supply Chain (25)
- PAT A 1079. Total Sales of Supply Chain (25)
- PAT 1079. Total Sales of Supply Chain (25)
- pat 1079. Total Sales of Supply Chain(树模拟)
- PAT 1079. Total Sales of Supply Chain (25)
- PAT 1079. Total Sales of Supply Chain (25)
- 【PAT】1079. Total Sales of Supply Chain (25)
- PAT 1014Total Sales of Supply Chain (25)
- 1079. Total Sales of Supply Chain (25)PAT甲级
- PAT A1079. Total Sales of Supply Chain (25)
- for循环遍历删除报错:java.util.ConcurrentModificationException
- Java 随机数
- hive中子查询实例
- 动画(五)属性动画的实现原理
- GOF共23种设计模式
- pat 1079 Total Sales of Supply Chain
- ListView random IndexOutOfBoundsException on Froyo
- 基于display:table的CSS布局让HTML元素和像table一样
- iOS面试题非技术面试(二)
- C++开发者都应该使用的10个C++11特性
- 在Android Studio添加SO library
- C++ 中的回调函数
- 动画(六)属性动画的工作原理
- Html5 本地存储