PAT (Advanced Level) Practise 1090 Highest Price in Supply Chain (25)

最新推荐文章于 2021-09-03 20:31:31 发布

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最新推荐文章于 2021-09-03 20:31:31 发布

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本文介绍了一种算法，用于计算供应链中从供应商到零售商的产品价格递增过程，并找出最高的零售价格及其出现次数。

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1090. Highest 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. 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 (<=10⁵), the total number of the members in the supply chain (and hence they are numbered from 0 to N-1); P, the price given by the root supplier; and r, the percentage rate of price increment for each distributor or retailer. Then the next line contains N numbers, each number S_i is the index of the supplier for the i-th member. S_root for the root supplier is defined to be -1. All the numbers in a line are separated by a space.

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 10¹⁰.

Sample Input:

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

Sample Output:

1.85 2

题意：一共有n个人，分别为零售商，经销商和供应商，供应商为根节点，零售商为叶子结点，给你一件物品的初始价钱，物品每经过一次传递会增加r%，问零售商最多能卖多少钱，有多少个供应商能卖最多的钱

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <cmath>
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <vector>
#include <bitset>

using namespace std;

#define LL long long
const int INF = 0x3f3f3f3f;

int n,a[100009],cnt,root;
double p,r,ma;
vector<int>g[100009];

void dfs(int k,double x)
{
    if(x>ma) ma=x,cnt=1;
    else if(x==ma) cnt++;
    int Size=g[k].size();
    for(int i=0;i<Size;i++)
        dfs(g[k][i],x*(1+1.0*r/100));
}

int main()
{
    while(~scanf("%d%lf%lf",&n,&p,&r))
    {
        ma=-1;
        for(int i=0;i<n;i++) g[i].clear();
        for(int i=0;i<n;i++)
        {
            scanf("%d",&a[i]);
            if(a[i]==-1) {root=i;continue;}
            g[a[i]].push_back(i);
        }
        dfs(root,p);
        printf("%.2lf %d\n",ma,cnt);
    }
    return 0;
}