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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.
输入描述:
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 Si is the index of the supplier for the i-th member. Sroot for the root supplier is defined to be -1. All the numbers in a line are separated by a space.
输出描述:
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.
16分做法:
思路:
直接用数组就行了，遍历每一个数组找到a[k]=-1，即就是找到根节点，然后用另一个数组记录每一个值，并维护最大值就可以了。当然会超时
#include<iostream>
#include<algorithm>
#include<algorithm>
using namespace std;
int n;
double p,r;
int a[100000];
int counts;
double maxn;
int k;
double nowcost[100000];
double currcost;
int main()
{cin>>n>>p>>r;
maxn=0;
r=1+0.01*r;for(int i=0;i<n;i++)cin>>a[i];for(int i=0;i<n;i++){k=i;currcost=p;while(a[k]!=-1)//找根的过程{k=a[k];
currcost*=r;if(a[k]==-1)break;}if(currcost>=maxn){maxn=currcost;}nowcost[i]=currcost;}for(int i=0;i<n;i++)if(nowcost[i]==maxn)counts++;printf("%.2lf",maxn);cout<<" ";
cout<<counts;return 0;
}
25分做法，bfs
vector二维数组数组模拟树搜下去就好了
#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int maxn=1e5+5;
int n,root,anscnt=0;double p,r;
vector<int> G[maxn];
map<int,int> mp;void dfs(int cur,int step){int ok=1;for(int i=0;i<G[cur].size();++i){int v=G[cur][i];ok=0;dfs(v,step+1); }if(ok){if(step>anscnt){anscnt=step;++mp[anscnt];return ;}if(step==anscnt){++mp[anscnt];return ;}}
}
int main(){scanf("%d %lf %lf",&n,&p,&r);for(int i=0,tmp;i<n;++i){scanf("%d",&tmp);if(tmp==-1){ root=i;continue;}G[tmp].push_back(i);}dfs(root,0); printf("%.2f %d\n",p*pow(1+r*0.01,anscnt),mp[anscnt]);return 0;
}
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