PAT Advanced 1090 Highest Price in Supply Chain: DFS or BFS

Overview

PAT 1090 Highest Price in Supply Chain solution: Do Tree construction, then use dfs or bfs to count the number of deepest leaf node and get the price.

PAT Advanced 1090 Highest Price in Supply Chain

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 (<=105), 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

Analysis: DFS or BFS

This problem is not difficult, first we could construct the whole tree by adjacent linked list, and we do dfs or bfs the get the number of deepest leaf node, suppose the height is h, then the price at that deep would be p * (1 + r / 100)^h. The following code is accepted by PAT OJ to pass this 1090 Highest Price in Supply Chain problem:

#include <cstdio>
#include <cmath>
#include <vector>

void dfs(const std::vector< std::vector<int> > &g, int root, int depth,
		 int *retailer, int *gDepth) {
	if (g[root].empty() ) {
		if (depth > *gDepth) {
			*retailer = 1;
			*gDepth = depth;
		}
		else if (depth == *gDepth) {
			++(*retailer);
		}
		return ;
	}
	for (auto it = g[root].begin(); it != g[root].end(); ++it) {
		dfs(g, *it, depth + 1, retailer, gDepth);
	}
}

void gao(int suppliers[], int n, double p, double r) {
	std::vector< std::vector<int> > g(n, std::vector<int>());
	int root = -1;
	for (int i = 0; i < n; ++i) {
		if (-1 == suppliers[i])
			root = i;
		else {
			g[suppliers[i]].push_back(i);
		}
	}
	int retailer = 0;
	int h = -1;
	dfs(g, root, 0, &retailer, &h);
	printf("%.2lf %d\n", p * std::pow(1 + r / 100.0, h), retailer);
}

int main() {
	int n(0);
	double p(0.0), r(0.0);
	scanf("%d %lf %lf", &n, &p, &r);
	int *suppliers = new int[n];
	for (int i = 0; i < n; ++i) scanf("%d", suppliers + i);
	gao(suppliers, n, p, r);
	delete [] suppliers;
	return 0;
}

Note do not put too many arguments in the recursive call dfs(), if there are too many arguments say 7 or 8 these many, then it would be much easier to get a segment fault error from the PAT OJ. From this fact, I assume PAT OJ do not allow too deep recursive calls.

Summary

PAT 1090 Highest Price in Supply Chain solution: Do Tree construction, then use dfs or bfs to count the number of deepest leaf node and get the price.

Written on December 11, 2014