PAT 1090 Highest Price in Supply Chain [DFS] [树的遍历]

最新推荐文章于 2022-10-09 18:03:48 发布

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最新推荐文章于 2022-10-09 18:03:48 发布

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本文探讨了供应链中从供应商到零售商的产品价格计算方法，通过构建供应链网络模型，使用深度优先搜索算法确定产品从源头到最终销售点的最大价格增长路径，并计算最高零售价及达到该价格的零售商数量。

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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.

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 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.

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

-----------------------------------这是题目和解题的分割线-----------------------------------

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. 可得，1是0的父结点，5是1的父结点，4是2的父结点，以此类推。题意是让求出最大层数，用dfs递归。

#include<cstdio>
#include<vector>
#include<cmath>

using namespace std;

vector<int> tree[100010]; //这棵树没有数据，只有子结点，所以不必定义结构体 
int n,last,root,maxL = 0,num = 0;
double money,rate,tmp; //注意用double别用float，不然结果有偏差 

//深搜找到最大层数 
void dfs(int index,int layer)
{
	if(tree[index].size()==0) //临界点，遍历到了叶子节点时 
	{
		if(layer>maxL)  
		{
			maxL = layer; //更新最大层数 
			num = 1; //并更新最大层数的路径数 
		}
		//如果和最大层数相等，路径数+1		
		else if(layer==maxL) num++; //不要else要else if，else还包括了小于的情况= =  
		return;
	}
	for(int i=0;i<tree[index].size();i++)
		dfs(tree[index][i],layer+1); //遍历子结点 
}

int main()
{
	scanf("%d%lf%lf",&n,&money,&rate);
	rate /= 100; //增长率 
	for(int i=0;i<n;i++)
	{
		scanf("%d",&last);
		if(last==-1) root = i; //父结点为-1的结点即为根节点 
		else tree[last].push_back(i);
	}
	dfs(root,0); //递归求出最大层数 
	printf("%.2f %d\n",money*pow(rate+1,maxL),num);
	return 0;
}

也可直接在DFS中算数

#include<cstdio>
#include<vector>

using namespace std;

struct node
{
	vector<int> child;
}tree[100010];

double maxS = 0,percent;
int n,root,cnt = 0;

void dfs(int index,double sum)
{
	if(tree[index].child.size()==0)
	{
		if(sum>maxS)
		{
			maxS = sum;
			cnt = 1;
		}
		else if(sum==maxS) cnt++;
		return;
	}
	if(index>n) return;
	for(int i=0;i<tree[index].child.size();i++)
	{
		int next = tree[index].child[i];
		dfs(next,sum+sum*percent/100);
		//dfs(next,sum);
	}
}

int main()
{
	int i;
	double begin;
	scanf("%d%lf%lf",&n,&begin,&percent);
	for(i=0;i<n;i++)
	{
		int x;
		scanf("%d",&x);
		if(x==-1) root = i;
		else tree[x].child.push_back(i);
	}
	dfs(root,begin);
	printf("%.2lf %d",maxS,cnt);
	return 0;
}