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1090 Highest Price in Supply Chain (25)(25 分)

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seo 发表于 2022-5-31 13:33:51 | 显示全部楼层 |阅读模式
1090 Highest Price in Supply Chain (25)(25 分)发布时间:2022/5/31 13:02:15
            
                                                       
                                                       
            
        
        
               
                     
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. 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 highest price we 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 (

Output Specification:

For each test case, print in one line the highest price we can expect from some retailers, accurate up to 2 decimal places, and the number of retailers that sell at the highest price. There must be one space between the two numbers. It is guaranteed that the price will not exceed 10^10^.

Sample Input:

9 1.80 1.00
1 5 4 4 -1 4 5 3 6

Sample Output:

1.85 2
  
  #include
#include
#include
using namespace std;
const int maxn = 100010;
vectorint> child[maxn];
int n,maxDepth = 0,num =0 ;
double p,r;
void DFS(int index,int depth){
    if(child[index].size() == 0){
        if(depth > maxDepth){
            maxDepth = depth;
            num = 1;
        }else if(depth == maxDepth){
            num++;
        }
        return;
    }
    for(int i = 0; i ){
        DFS(child[index],depth+1);
    }
}
int main(){
    scanf("%d%lf%lf",&n,&p,&r);
    int father,root;
    r /= 100;
    for(int i = 0; i ){
        scanf("%d",&father);
        if(father != -1){
            child[father].push_back(i);
        }
        else{
            root = i;
        }
    }
    DFS(root,0);
    printf("%.2f %d\n",p*pow(1+r,maxDepth),num);
    return 0;
}
  
  


转载于:https://www.cnblogs.com/wanghao-boke/p/9291501.html
               
        
        
   
            
        
        
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