PAT甲级1106
来源:互联网 发布:linux程序设计pdf下载 编辑:程序博客网 时间:2024/05/01 12:11
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 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 (<=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 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 1010.
Sample Input:10 1.80 1.003 2 3 51 91 41 702 6 11 8000Sample Output:
1.8362 2
#include<cstdio>#include<vector>#include<algorithm>using namespace std;int N;double P, r;const int maxn = 100000;vector<int> children[maxn];int minlevel = 1000000000;void DFS(int root, int l){if (!children[root].size()){minlevel = min(l, minlevel);return;}for (int i = 0; i < children[root].size(); i++){DFS(children[root][i], l + 1);}}void DFS(int root, int l, int &count){if (!children[root].size()){if (l == minlevel)count++;return;}for (int i = 0; i < children[root].size(); i++){DFS(children[root][i], l + 1, count);}}int main(){scanf("%d%lf%lf", &N, &P, &r);int t,child;for (int i = 0; i < N; i++){scanf("%d", &t);for (int j = 0; j < t; j++){scanf("%d", &child);children[i].push_back(child);} }DFS(0, 0);int count = 0;DFS(0, 0, count);printf("%.4f %d", P*pow((1.0 + r / 100), minlevel), count);return 0;}
- PAT甲级1106
- PAT 甲级
- 浙大PAT甲级 1106 广度优先搜索
- PAT甲级 A1025.PAT RANKING
- PAT 甲级 1025 PAT Ranking
- PAT(甲级)1003
- PAT(甲级)1004
- PAT(甲级)1005
- PAT(甲级)1006
- PAT(甲级)1007
- PAT(甲级)1008
- PAT(甲级)1009
- PAT(甲级)1010
- PAT(甲级)1011
- PAT(甲级)1012
- PAT(甲级)1013
- PAT(甲级)1014
- PAT(甲级)1015
- 一个月学习C++笔记(四)
- 数据库异常整理:org.hibernate.QueryException: could not resolve property: “xxx”
- VMware 虚拟上网的的三种模式 ——bridged、host-only、NAT 模式--学习整理笔记
- leecode 解题总结:17. Letter Combinations of a Phone Number
- [Kibana]如何支持携带身份信息的请求到Elasticsearch服务器
- PAT甲级1106
- Ubuntn16.04下安装和配置arm-linux-gcc
- javascript实现-最简单选项卡切换
- iOS开发-SQLite数据库在App项目中的设置及使用
- java rmi 初探
- LeetCode 21 Merge Two Sorted Lists
- 灾后重建_洛谷1119_最短路
- 存储器这个小话题(3)
- ubuntu 16.04安装搜狗输入法