PAT A1106 Lowest Price in Supply Chain (25point(s))

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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 (≤$10^5$), 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:KiID[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

题意

寻找最短供应链,打印出该供应链的最终价格,和供应链的条数。

解答

使用dfs算法即可,可以剪枝,即当前深度已经超过最小深度,就可以结束,而且不需要visited数组,因为构造的是有向图,而且是单向的。本题使用scanf代替cin可以加快执行速度,使用cin需要100ms左右,而scanf只需要30ms。

#define _CRT_SECURE_NO_WARNINGS
#include<iostream>
#include<vector>
#include<cmath>

using namespace std;

int count1 = 0, n;
int minDepth = 100000;
vector<vector<int>> adj;

void dfs(int id, int depth)
{
	if (depth > minDepth) return;
	bool flag = false;
	for (int i = 0; i < adj[id].size(); i++) {
		if (adj[id][i]) {
			flag = true;
			dfs(adj[id][i], depth + 1);
		}
	}
	if (!flag && depth <= minDepth) {
		if (depth == minDepth) count1++;
		else {
			count1 = 1;
			minDepth = depth;
		}
	}
}

int main()
{
	int memNum, mem;
	double rootPrice, rate;
	scanf("%d %lf %lf", &n, &rootPrice, &rate);
	adj = vector<vector<int>>(n);
	for(int i =0;i<n;i++){
		scanf("%d", &memNum);
		for (int j = 0; j < memNum; j++) {
			scanf("%d", &mem);
			adj[i].push_back(mem);
		}
	}
	dfs(0, 0);
	printf("%.4f %d", rootPrice*pow(rate / 100.0 + 1.0, minDepth), count1);
	return 0;
}