solution Of 1106. Lowest Price in Supply Chain (25)

1106. Lowest Price in Supply Chain (25)

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

结题思路 :

题意要求我们计算找到可以后的最低价格的零售商的价格和这个价格的零售商的个数。

要求1：每个中间商或零售商只有一个上游卖家；

要点2：index=0作为生产商；

要点3：供应链中不存在环路。

程序步骤：

第一步、初始化，建立索引关系；

第二步、对每个没有下游买家的零售商求取他们距离生产商经过了几个中间商。

第三部、找到零售商的集合中距离生产商最近的零售商个数。

具体程序(AC)如下：

#include <iostream>
#include <cstring>
#include <string>
#include <vector>
#include <cmath>
using namespace std;
int parent[100005];
int depth[100005];
vector<int> retailers;
int getDepth(int site)
{
if(site==0)
return 0;
int pa=parent[site];
if(depth[pa]!=-1)
return depth[site]=depth[pa]+1;
return depth[site]=getDepth(pa)+1;
}
int main()
{
int num;
double priceRoot;
double rate;
cin>>num>>priceRoot>>rate;
memset(parent,-1,sizeof(parent));
memset(depth,-1,sizeof(depth));
depth[0]=0;//root
retailers.clear();
int childNum,child;
for(int i=0;i<num;++i)
{
cin>>childNum;
if(childNum==0)
retailers.push_back(i);
for(int j=0;j<childNum;++j)
{
cin>>child;
parent[child]=i;
}
}
int min=100005;
int availab=0;
for(int i=0;i<retailers.size();++i)
{
depth[retailers[i]]=getDepth(retailers[i]);
if(depth[retailers[i]]<min)
{
min=depth[retailers[i]];
availab=1;
}
else if(depth[retailers[i]]==min)
++availab;
}
rate=pow(1+rate/100,min);
printf("%.4lf %d\n",priceRoot*rate,availab);
}