1106 Lowest Price in Supply Chain（25 分）(cj)

最新推荐文章于 2020-03-30 19:41:58 发布

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本文探讨了从供应商到消费者的供应链网络中，如何计算最低价格。通过分析每个成员从供应商处购买产品并以一定百分比加价出售的过程，我们提供了一个算法来找出消费者能期待的最低价格。

1106 Lowest Price in Supply Chain（25 分）

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 1.00
3 2 3 5
1 9
1 4
1 7
0
2 6 1
1 8
0
0
0

Sample Output:

1.8362 2

又是供应商，零售商，经销商的爱恨情仇。。。

code

#pragma warning(disable:4996)
#include <iostream>
#include <vector>
#include <stdio.h>
using namespace std;
vector<int> varr[100005];
int n, num = 0;;
double price, rate;
double minprice = 99999999;
void dfs(int pos, double price);
int main() {
	cin >> n >> price >> rate;
	int k, x;
	for (int i = 0; i < n; ++i) {
		cin >> k;
		while (k--) {
			cin >> x;
			varr[i].push_back(x);
		}
	}
	dfs(0, price);
	printf("%.4f %d", minprice, num);
	system("pause");
	return 0;
}
void dfs(int pos, double tprice) {
	if (varr[pos].size() == 0) {
		if (tprice < minprice) {
			minprice = tprice;
			num = 1;
		}
		else if (tprice == minprice) num++;
	}
	for (int i = 0; i < varr[pos].size(); ++i) {
		dfs(varr[pos][i], tprice*(1.0 + rate / 100));
	}
}