PAT-PAT (Advanced Level) Practise Highest Price in Supply Chain(25) 【三星级】

题目链接：http://www.patest.cn/contests/pat-a-practise/1090

题面：

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 1.001 5 4 4 -1 4 5 3 6

Sample Output:

1.85 2

题目大意：

   -1是根节点，其他的是位置的值，是其父亲节点的位置值。下标从0~N-1,求原值p乘以(1+0.01*r)^x的值，和最后一层的节点数。x为层深。

代码：

   

#include <iostream>#include <cstdio>#include <algorithm>#include <queue>#include <vector>using namespace std;int ori[100010],depth[100010];vector <int> v[100010];queue <int> qe;int main(){int n,cnt=1,pos,tmp,dep;double p,r;//读入 scanf("%d%lf%lf",&n,&p,&r);for(int i=0;i<n;i++){    scanf("%d",&ori[i]);    if(ori[i]==-1)      pos=i;}//每个父亲节点记录其儿子节点 for(int i=0;i<n;i++){v[ori[i]].push_back(i);}depth[pos]=1;qe.push(pos);//队列模拟遍历树 while(!qe.empty()){tmp=qe.front();dep=depth[tmp]+1;qe.pop();for(int i=0;i<v[tmp].size();i++){depth[v[tmp][i]]=dep;qe.push(v[tmp][i]);} }//排个序 sort(depth,depth+n);dep=depth[n-1];for(int i=n-2;i>=0;i--){if(depth[i]==dep)  cnt++;        else break;} //求结果 for(int i=1;i<dep;i++)  p*=(1+0.01*r);printf("%.2lf %d\n",p,cnt);return 0;}