1090. Highest Price in Supply Chain (25)【树】——PAT (Advanced Level) Practise
来源:互联网 发布:淘宝喵喵通信 编辑:程序博客网 时间:2024/04/20 21:46
题目信息
1090. Highest Price in Supply Chain (25)
时间限制200 ms
内存限制65536 kB
代码长度限制16000 B
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. 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 highest price we 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 they are numbered from 0 to N-1); P, the price given by the root supplier; and r, the percentage rate of price increment for each distributor or retailer. Then the next line contains N numbers, each number Si is the index of the supplier for the i-th member. Sroot for the root supplier is defined to be -1. All the numbers in a line are separated by a space.
Output Specification:
For each test case, print in one line the highest price we can expect from some retailers, accurate up to 2 decimal places, and the number of retailers that sell at the highest price. There must be one space between the two numbers. It is guaranteed that the price will not exceed 1010.
Sample Input:
9 1.80 1.00
1 5 4 4 -1 4 5 3 6
Sample Output:
1.85 2
解题思路
建树搜索
AC代码
#include <cstdio>#include <vector>#include <cmath>using namespace std;int fa[100005];vector<int> sroot, child[100005];int maxLevel = -1, maxNum;void dfs(int root, int lv){ if (lv > maxLevel){ maxLevel = lv; maxNum = 1; }else if (lv == maxLevel){ ++maxNum; } for (int i = 0; i < child[root].size(); ++i){ dfs(child[root][i], lv + 1); }}int main(){ int n, t; double p, r; scanf("%d%lf%lf", &n, &p, &r); for (int i = 0; i < n; ++i){ scanf("%d", &t); fa[i] = t; if (t == -1){ sroot.push_back(i); }else{ child[t].push_back(i); } } int mx = 0; for (int i = 0; i < sroot.size(); ++i){ dfs(sroot[i], 0); } printf("%.2f %d\n", p * pow(1+r/100, maxLevel), maxNum); return 0;}
- 1090. Highest Price in Supply Chain (25)【树】——PAT (Advanced Level) Practise
- 【PAT】【Advanced Level】1090. Highest Price in Supply Chain (25)
- PAT-PAT (Advanced Level) Practise Highest Price in Supply Chain(25) 【三星级】
- 1106. Lowest Price in Supply Chain (25)【树+深搜】——PAT (Advanced Level) Practise
- PAT (Advanced Level) Practise 1090 Highest Price in Supply Chain (25)
- PAT (Advanced Level) Practise 1090 Highest Price in Supply Chain (25)
- PAT (Advanced Level) 1090. Highest Price in Supply Chain (25) 供应链,BFS
- Pat(Advanced Level)Practice--1090(Highest Price in Supply Chain)
- PAT (Advanced Level) Practise 1106 Lowest Price in Supply Chain (25)
- PAT (Advanced Level) Practise 1106 Lowest Price in Supply Chain(25)
- PAT 1090. Highest Price in Supply Chain
- 【PAT】1090. Highest Price in Supply Chain
- 【PAT】【Advanced Level】1106. Lowest Price in Supply Chain (25)
- PAT 1090. Highest Price in Supply Chain (25)
- PAT A 1090. Highest Price in Supply Chain (25)
- PAT 1090. Highest Price in Supply Chain (25)(DFS)
- PAT 1090. Highest Price in Supply Chain (25)
- 【PAT】1090. Highest Price in Supply Chain (25)
- 1089. Insert or Merge (25)【排序】——PAT (Advanced Level) Practise
- linux centos lamp开启GD库的支持
- "围观"设计模式(10)--创建型之原型模式(Prototype Pattern)
- 微信JS SDK PHP Demo
- 分布式环境Session处理方法
- 1090. Highest Price in Supply Chain (25)【树】——PAT (Advanced Level) Practise
- int与Integer的区别
- php &$
- bestcoder百度之星2016AK 1001&1002&1003&1004 题解
- Windows网络编程学习笔记(4) 编写一个TCP客户端
- HYSBZ 1036树的统计Count 树链剖分
- PHP中的可变参数函数和可选参数函数
- 1091. Acute Stroke (30)【搜索】——PAT (Advanced Level) Practise
- 警察与厨师