[PAT A1106]Lowest Price in Supply Chain

[PAT A1106]Lowest Price in Supply Chain

题目描述

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.

输入格式

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:

K?i?? ID[1] ID[2] ... ID[K?i??]

where in the i-th line, K?i?? 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. K?j?? being 0 means that the j-th member is a retailer. All the numbers in a line are separated by a space.

输出格式

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 ???.

输入样例

10 1.80 1.00
3 2 3 5
1 9
1 4
1 7
0
2 6 1
1 8
0
0
0

输出样例

1.8362 2

解析

1.题目大意：使用的是A1079的输入格式和A1090的输出格式，具体题意不再阐述

参考地址：

[PAT A1079]Total Sales of Supply Chain

[PAT A1090]Highest Price in Supply Chain

2.代码如下：

#include<iostream>
#include<vector>
#include<cmath>   //计算次方的pow函数，我们需要使用cmath头文件
const int maxn = 100010;  //不超过10^5个结点
using namespace std;
vector<int> tree[maxn];  //定义数据结构
bool notroot[maxn] = { false }; //寻找根结点的辅助数组
int cnt[maxn] = { 0 };       //记录每层叶结点的个数
void traverse(int i, int level);  //遍历函数
int minl = maxn;
int main()
{int n;double p, r;scanf("%d %lf %lf", &n, &p, &r);for (int i = 0; i < n; i++) {  //用于输入数据和填写根结点数组（判断根结点位置）int t, temp;scanf("%d", &t);for (int j = 0; j < t; j++) {scanf("%d", &temp);notroot[temp] = true;tree[i].push_back(temp);}}int root = -1;for (int i = 0; i < n; i++) {  //寻找根结点if (notroot[i] == false) {root = i;break;}}traverse(root, 0);printf("%.4f %d", p*pow(r / 100 + 1, minl), cnt[minl]);return 0;
}
void traverse(int i, int level) {if (tree[i].empty()) {cnt[level]++;if (level < minl) minl = level;return;}for (int j = 0; j < tree[i].size(); j++)traverse(tree[i][j], level + 1);
}