1106. Lowest Price in Supply Chain (25)

时间限制

200 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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:

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.

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 1.003 2 3 51 91 41 702 6 11 8000

Sample Output:

1.8362 2

这一题的类型感觉去年做了好几次了。但是这次还是会细节问题

N（三种商人）  price（厂家价格）  r（每转手一家，价格（1+r%））

ki个  ID[1]……ID[ki]

[其中ID从0到N-1，ID[]=0代表厂家]

问题求的是没有转手的那些商家【零售商】，价格最低是多少【min】，且达到最低价格的有几家【num】

输出

min  num

评测结果

时间结果得分题目语言用时(ms)内存(kB)用户7月08日 16:39答案正确251106C++ (g++ 4.7.2)805880datrilla

测试点

测试点结果用时(ms)内存(kB)得分/满分0答案正确237211/111答案正确23841/12答案正确7230442/23答案正确8058802/24答案正确43843/35答案正确7643402/26答案正确8044802/27答案正确8058562/2

#include<iostream>#include<vector>#include<queue>#include<iomanip>using namespace std;void redln(vector<vector<int > >*retailers,int N){   for(int i=0;i<N;i++)  {    int m=0;    cin>>m;    (*retailers)[i].resize(m);    for(int j=0;j<m;j++)    {      int t;      cin>>t;      (*retailers)[i][j]=t;     }  }}void bfs_(vector<vector<int > >*retailers,double root_price,double r,double *min,int*num){   queue<int >qt;   if(0==(*retailers)[0].size())   {     (*min)=root_price;       (*num)=1;   }else   qt.push(0);   while(!qt.empty())   {      int len=qt.size();     root_price*=(1+r/100);      for(int i=0;i<len;i++)     {            for(int add=0;add<(*retailers)[qt.front()].size();add++)        {          int tid=(*retailers)[qt.front()][add];                  if(0==(*retailers)[tid].size())          {                 if(0==(*num)||(*min)>root_price)            {                  (*min)=root_price;                          (*num)=1;            }                  else if((*num)>0&&(*min)==root_price)                (*num)++;           }          else          qt.push(tid);        }        qt.pop();     }   }}int main(){  int N;  double root_price,r;  cin>>N>>root_price>>r;  vector<vector<int > >retailers;  retailers.resize(N);  redln(&retailers,N);   double min=root_price;  int num=0;  bfs_(&retailers,root_price,r,&min,&num);  cout<<setiosflags(ios::fixed)<<setprecision(4)<<min<<" "<<num<<endl;    return 0;}